Exercise 3.1 - Question 1

Question: Find the radian measures corresponding to the following degree measures: (i) $25^\circ$ (ii) $-47^\circ 30'$ (iii) $240^\circ$ (iv) $520^\circ$

Solution: We know that $180^\circ = \pi \text{ radians}$, so $1^\circ = \frac{\pi}{180} \text{ radians}$. To convert a degree measure into radian measure, we multiply the given degree measure by $\frac{\pi}{180}$.

(i) $25^\circ$ $$\text{Radian measure} = 25 \times \frac{\pi}{180} = \frac{5\pi}{36}$$
(ii) $-47^\circ 30'$ First, convert $30'$ into degrees: $$30' = \left(\frac{30}{60}\right)^\circ = \left(\frac{1}{2}\right)^\circ = 0.5^\circ$$ Thus, $-47^\circ 30' = -47.5^\circ = -\frac{95}{2}^\circ$. $$\text{Radian measure} = -\frac{95}{2} \times \frac{\pi}{180} = -\frac{19}{2} \times \frac{\pi}{36} = -\frac{19\pi}{72}$$
(iii) $240^\circ$ $$\text{Radian measure} = 240 \times \frac{\pi}{180} = \frac{24\pi}{18} = \frac{4\pi}{3}$$
(iv) $520^\circ$ $$\text{Radian measure} = 520 \times \frac{\pi}{180} = \frac{52\pi}{18} = \frac{26\pi}{9}$$

Exercise 3.1 - Question 2

Question: Find the degree measures corresponding to the following radian measures (Use $\pi = \frac{22}{7}$): (i) $\frac{11}{16}$ (ii) $-4$ (iii) $\frac{5\pi}{3}$ (iv) $\frac{7\pi}{6}$

Solution: We know that $\pi \text{ radians} = 180^\circ$, so $1 \text{ radian} = \frac{180}{\pi} \text{ degrees}$. To convert a radian measure into degree measure, we multiply by $\frac{180}{\pi}$ (using $\pi = \frac{22}{7}$).

(i) $\frac{11}{16}$ $$\text{Degree measure} = \frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22} = \frac{180 \times 7}{16 \times 2} = \frac{315}{8}^\circ$$ $$\frac{315}{8} = 39^\circ \text{ with remainder } \frac{3}{8}^\circ$$ $$\frac{3}{8} \times 60' = \frac{180}{8}' = 22.5' = 22' + 0.5'$$ $$0.5' = 0.5 \times 60'' = 30''$$ Thus, $39^\circ 22' 30''$.
(ii) $-4$ $$\text{Degree measure} = -4 \times \frac{180}{\pi} = -4 \times \frac{180 \times 7}{22} = -\frac{2 \times 180 \times 7}{11} = -\frac{2520}{11}^\circ$$ $$-\frac{2520}{11} = -229\frac{1}{11}^\circ = -229^\circ - \left(\frac{1}{11} \times 60\right)' = -229^\circ - \left(\frac{60}{11}\right)'$$ $$-\left(5\frac{5}{11}\right)' = -229^\circ 5' - \left(\frac{5}{11} \times 60\right)'' = -229^\circ 5' - \left(\frac{300}{11}\right)''$$ $$\approx -229^\circ 5' 27''$$
(iii) $\frac{5\pi}{3}$ $$\text{Degree measure} = \frac{5\pi}{3} \times \frac{180}{\pi} = \frac{5 \times 180}{3} = 5 \times 60^\circ = 300^\circ$$
(iv) $\frac{7\pi}{6}$ $$\text{Degree measure} = \frac{7\pi}{6} \times \frac{180}{\pi} = \frac{7 \times 180}{6} = 7 \times 30^\circ = 210^\circ$$

Exercise 3.1 - Question 3

Question: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution: In one complete revolution, a wheel turns through an angle of $2\pi \text{ radians}$.
Number of revolutions made by the wheel in 1 minute ($60 \text{ seconds}$) = $360$.
Number of revolutions made by the wheel in 1 second = $\frac{360}{60} = 6$.
Therefore, the angle turned through in 1 second = $6 \times 2\pi \text{ radians} = 12\pi \text{ radians}$.

Exercise 3.1 - Question 4

Question: Find the degree measure of the angle subtended at the centre of a circle of radius $100\text{ cm}$ by an arc of length $22\text{ cm}$ (Use $\pi = \frac{22}{7}$).

Solution: We know that the relation between the radius $r$, the arc length $l$, and the angle $\theta$ (in radians) subtended at the centre is given by: $$\theta = \frac{l}{r}$$ Given: - Radius of the circle, $r = 100\text{ cm}$ - Length of the arc, $l = 22\text{ cm}$ Substitute the given values into the formula: $$\theta = \frac{22}{100}\text{ radians} = \frac{11}{50}\text{ radians}$$ To convert the angle from radians to degrees, we multiply by $\frac{180^\circ}{\pi}$ (using $\pi = \frac{22}{7}$): $$\theta^\circ = \frac{11}{50} \times \frac{180}{\pi} = \frac{11}{50} \times \frac{180 \times 7}{22}$$ Simplify the expression: $$\theta^\circ = \frac{11 \times 180 \times 7}{50 \times 22} = \frac{180 \times 7}{50 \times 2} = \frac{1260}{100} = 12.6^\circ$$ Converting $0.6^\circ$ into minutes: $$0.6^\circ = 0.6 \times 60' = 36'$$ Thus, the degree measure of the angle is: $$12^\circ 36'$$

Exercise 3.1 - Question 5

Question: In a circle of diameter $40\text{ cm}$, the length of a chord is $20\text{ cm}$. Find the length of minor arc of the chord.

Solution: Given: - Diameter of the circle = $40\text{ cm}$ - Radius of the circle, $r = \frac{40}{2} = 20\text{ cm}$ - Length of the chord, $AB = 20\text{ cm}$ Let $O$ be the centre of the circle. Join $OA$ and $OB$. In $\triangle OAB$: - $OA = r = 20\text{ cm}$ (radius) - $OB = r = 20\text{ cm}$ (radius) - $AB = 20\text{ cm}$ (chord) Since all three sides of $\triangle OAB$ are equal ($OA = OB = AB = 20\text{ cm}$), $\triangle OAB$ is an equilateral triangle. Therefore, the angle subtended by the chord at the centre, $\theta$, is: $$\theta = 60^\circ = 60 \times \frac{\pi}{180}\text{ radians} = \frac{\pi}{3}\text{ radians}$$ The length of the minor arc $l$ is given by the formula: $$l = r \theta$$ Substitute the values of $r$ and $\theta$: $$l = 20 \times \frac{\pi}{3} = \frac{20\pi}{3}\text{ cm}$$ Thus, the length of the minor arc is $\frac{20\pi}{3}\text{ cm}$.

Exercise 3.1 - Question 6

Question: If in two circles, arcs of the same length subtend angles $60^\circ$ and $75^\circ$ at the centre, find the ratio of their radii.

Solution: Let the radii of the two circles be $r_1$ and $r_2$, and let the length of the arc in both circles be $l$. The angles subtended by the arcs at the centres are: - $\theta_1 = 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}\text{ radians}$ - $\theta_2 = 75^\circ = 75 \times \frac{\pi}{180} = \frac{5\pi}{12}\text{ radians}$ We know that $l = r\theta$, which means for the two circles: $$l = r_1 \theta_1 \implies r_1 = \frac{l}{\theta_1}$$ $$l = r_2 \theta_2 \implies r_2 = \frac{l}{\theta_2}$$ Now, find the ratio of their radii $r_1 : r_2$: $$\frac{r_1}{r_2} = \frac{\frac{l}{\theta_1}}{\frac{l}{\theta_2}} = \frac{\theta_2}{\theta_1}$$ Substitute the radian values of $\theta_1$ and $\theta_2$: $$\frac{r_1}{r_2} = \frac{\frac{5\pi}{12}}{\frac{\pi}{3}} = \frac{5\pi}{12} \times \frac{3}{\pi} = \frac{15}{12} = \frac{5}{4}$$ Thus, the ratio of their radii is $5 : 4$.

Exercise 3.1 - Question 7

Question: Find the angle in radian through which a pendulum swings if its length is $75\text{ cm}$ and the tip describes an arc of length (i) $10\text{ cm}$ (ii) $15\text{ cm}$ (iii) $21\text{ cm}$

Solution: The length of the pendulum acts as the radius of the circle, $r = 75\text{ cm}$. The angle $\theta$ in radians is given by $\theta = \frac{l}{r}$, where $l$ is the arc length described by the tip. **(i) When $l = 10\text{ cm}$:** $$\theta = \frac{10}{75} = \frac{2}{15}\text{ radians}$$ **(ii) When $l = 15\text{ cm}$:** $$\theta = \frac{15}{75} = \frac{1}{5}\text{ radians}$$ **(iii) When $l = 21\text{ cm}$:** $$\theta = \frac{21}{75} = \frac{7}{25}\text{ radians}$$

Exercise 3.2 - Question 1

Question: $\cos x = -\frac{1}{2}$, $x$ lies in third quadrant.

Solution:

Given that $\cos x = -\frac{1}{2}$ and $x$ lies in the third quadrant.

We know that $\sin^2 x + \cos^2 x = 1$. Substituting the value of $\cos x$:

$$\sin^2 x + \left(-\frac{1}{2}\right)^2 = 1$$

$$\sin^2 x + \frac{1}{4} = 1$$

$$\sin^2 x = 1 - \frac{1}{4} = \frac{3}{4}$$

Since $x$ lies in the third quadrant, $\sin x$ must be negative. Therefore:

$$\sin x = -\frac{\sqrt{3}}{2}$$

Now, we find $\text{cosec } x$:

$$\text{cosec } x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$$

Next, we find $\tan x$ using the relation $\tan x = \frac{\sin x}{\cos x}$:

$$\tan x = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$$

Now, we find $\cot x$:

$$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$$

Finally, we find $\sec x$:

$$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{1}{2}} = -2$$

Answer:
$\sin x = -\frac{\sqrt{3}}{2}$, $\text{cosec } x = -\frac{2}{\sqrt{3}}$, $\sec x = -2$, $\tan x = \sqrt{3}$, $\cot x = \frac{1}{\sqrt{3}}$

Exercise 3.2 - Question 2

Question: $\sin x = \frac{3}{5}$, $x$ lies in second quadrant.

Solution:

Given that $\sin x = \frac{3}{5}$ and $x$ lies in the second quadrant.

We know that $\sin^2 x + \cos^2 x = 1$. Substituting the value of $\sin x$:

$$\left(\frac{3}{5}\right)^2 + \cos^2 x = 1$$

$$\frac{9}{25} + \cos^2 x = 1$$

$$\cos^2 x = 1 - \frac{9}{25} = \frac{16}{25}$$

Since $x$ lies in the second quadrant, $\cos x$ must be negative. Therefore:

$$\cos x = -\frac{4}{5}$$

Now, we find $\text{cosec } x$:

$$\text{cosec } x = \frac{1}{\sin x} = \frac{5}{3}$$

Next, we find $\sec x$:

$$\sec x = \frac{1}{\cos x} = -\frac{5}{4}$$

Now, we find $\tan x$ using the relation $\tan x = \frac{\sin x}{\cos x}$:

$$\tan x = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$$

Finally, we find $\cot x$:

$$\cot x = \frac{1}{\tan x} = -\frac{4}{3}$$

Answer:
$\text{cosec } x = \frac{5}{3}$, $\cos x = -\frac{4}{5}$, $\sec x = -\frac{5}{4}$, $\tan x = -\frac{3}{4}$, $\cot x = -\frac{4}{3}$

Exercise 3.2 - Question 3

Question: $\cot x = \frac{3}{4}$, $x$ lies in third quadrant.

Solution:

Given that $\cot x = \frac{3}{4}$ and $x$ lies in the third quadrant.

First, we find $\tan x$:

$$\tan x = \frac{1}{\cot x} = \frac{4}{3}$$

We know the identity $1 + \tan^2 x = \sec^2 x$:

$$\sec^2 x = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$

Since $x$ lies in the third quadrant, $\sec x$ must be negative. Therefore:

$$\sec x = -\frac{5}{3}$$

Now, we find $\cos x$:

$$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$$

Next, we find $\sin x$ using the relation $\sin x = \tan x \cdot \cos x$:

$$\sin x = \left(\frac{4}{3}\right) \cdot \left(-\frac{3}{5}\right) = -\frac{4}{5}$$

Finally, we find $\text{cosec } x$:

$$\text{cosec } x = \frac{1}{\sin x} = -\frac{5}{4}$$

Answer:
$\sin x = -\frac{4}{5}$, $\text{cosec } x = -\frac{5}{4}$, $\cos x = -\frac{3}{5}$, $\sec x = -\frac{5}{3}$, $\tan x = \frac{4}{3}$

Exercise 3.2 - Question 4

Question: $\sec x = \frac{13}{5}$, $x$ lies in fourth quadrant.

Solution:

Given that $\sec x = \frac{13}{5}$ and $x$ lies in the fourth quadrant.

First, we find $\cos x$:

$$\cos x = \frac{1}{\sec x} = \frac{5}{13}$$

We know that $\sin^2 x + \cos^2 x = 1$. Substituting the value of $\cos x$:

$$\sin^2 x + \left(\frac{5}{13}\right)^2 = 1$$

$$\sin^2 x + \frac{25}{169} = 1$$

$$\sin^2 x = 1 - \frac{25}{169} = \frac{144}{169}$$

Since $x$ lies in the fourth quadrant, $\sin x$ must be negative. Therefore:

$$\sin x = -\frac{12}{13}$$

Now, we find $\text{cosec } x$:

$$\text{cosec } x = \frac{1}{\sin x} = -\frac{13}{12}$$

Next, we find $\tan x$ using the relation $\tan x = \frac{\sin x}{\cos x}$:

$$\tan x = \frac{-\frac{12}{13}}{\frac{5}{13}} = -\frac{12}{5}$$

Finally, we find $\cot x$:

$$\cot x = \frac{1}{\tan x} = -\frac{5}{12}$$

Answer:
$\sin x = -\frac{12}{13}$, $\text{cosec } x = -\frac{13}{12}$, $\cos x = \frac{5}{13}$, $\tan x = -\frac{12}{5}$, $\cot x = -\frac{5}{12}$

Exercise 3.2 - Question 5

Question: $\tan x = -\frac{5}{12}$, $x$ lies in second quadrant.

Solution:

Given that $\tan x = -\frac{5}{12}$ and $x$ lies in the second quadrant. In the second quadrant, sine and cosecant are positive, while cosine, secant, tangent, and cotangent are negative.

First, we find $\cot x$, which is the reciprocal of $\tan x$:

$$\cot x = \frac{1}{\tan x} = \frac{1}{-\frac{5}{12}} = -\frac{12}{5}$$

Next, we use the identity $1 + \tan^2 x = \sec^2 x$ to find $\sec x$:

$$\sec^2 x = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$$

Since $x$ is in the second quadrant, $\sec x$ is negative:

$$\sec x = -\sqrt{\frac{169}{144}} = -\frac{13}{12}$$

Now, $\cos x$ is the reciprocal of $\sec x$:

$$\cos x = \frac{1}{\sec x} = -\frac{12}{13}$$

Using the relation $\tan x = \frac{\sin x}{\cos x}$, we can find $\sin x$:

$$\sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right) \cdot \left(-\frac{12}{13}\right) = \frac{5}{13}$$

Finally, $\operatorname{cosec} x$ is the reciprocal of $\sin x$:

$$\operatorname{cosec} x = \frac{1}{\sin x} = \frac{13}{5}$$

Final Answer:
$\sin x = \frac{5}{13}$, $\operatorname{cosec} x = \frac{13}{5}$, $\cos x = -\frac{12}{13}$, $\sec x = -\frac{13}{12}$, $\cot x = -\frac{12}{5}$

Exercise 3.2 - Question 6

Question: $\sin 765^\circ$

Solution:

We know that the values of trigonometric functions repeat after an interval of $360^\circ$ (or $2\pi$). We can express $765^\circ$ as a multiple of $360^\circ$ plus a remainder:

$$765^\circ = 2 \times 360^\circ + 45^\circ$$ $$765^\circ = 720^\circ + 45^\circ$$

Using the periodicity property $\sin(n \cdot 360^\circ + \theta) = \sin \theta$ for any integer $n$, we get:

$$\sin(765^\circ) = \sin(2 \times 360^\circ + 45^\circ) = \sin(45^\circ)$$

From standard trigonometric values:

$$\sin(45^\circ) = \frac{1}{\sqrt{2}}$$

Final Answer:
$\frac{1}{\sqrt{2}}$

Exercise 3.2 - Question 7

Question: $\operatorname{cosec} (-1410^\circ)$

Solution:

First, use the negative angle property for cosecant, $\operatorname{cosec}(-\theta) = -\operatorname{cosec}(\theta)$:

$$\operatorname{cosec}(-1410^\circ) = -\operatorname{cosec}(1410^\circ)$$

Next, find the equivalent angle by dividing $1410^\circ$ by $360^\circ$:

$$1410^\circ = 3 \times 360^\circ + 330^\circ$$ $$1410^\circ = 1080^\circ + 330^\circ$$

Thus,

$$\operatorname{cosec}(1410^\circ) = \operatorname{cosec}(3 \times 360^\circ + 330^\circ) = \operatorname{cosec}(330^\circ)$$

Now, express $330^\circ$ using $360^\circ$:

$$\operatorname{cosec}(330^\circ) = \operatorname{cosec}(360^\circ - 30^\circ)$$

Since $\operatorname{cosec}(360^\circ - \theta) = -\operatorname{cosec}(\theta)$ (angle lies in the fourth quadrant where cosecant is negative):

$$\operatorname{cosec}(360^\circ - 30^\circ) = -\operatorname{cosec}(30^\circ)$$

Substitute this back into our original expression:

$$-\operatorname{cosec}(1410^\circ) = -(-\operatorname{cosec}(30^\circ)) = \operatorname{cosec}(30^\circ)$$

Since $\operatorname{cosec}(30^\circ) = 2$:

Final Answer:
$2$

Exercise 3.2 - Question 8

Question: $\tan \frac{19\pi}{3}$

Solution:

We divide the numerator by the denominator to find how many full periods of $2\pi$ are contained in $\frac{19\pi}{3}$:

$$\frac{19\pi}{3} = 6\pi + \frac{\pi}{3} = 3(2\pi) + \frac{\pi}{3}$$

Since the tangent function has a period of $\pi$ (and thus $2\pi$ as well), we have $\tan(2n\pi + \theta) = \tan \theta$:

$$\tan\left(\frac{19\pi}{3}\right) = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right)$$

From standard trigonometric values:

$$\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$

Final Answer:
$\sqrt{3}$

Trigonometric Functions - Question 9

Question: $\sin \left(-\frac{11\pi}{3}\right)$

Solution: We know that $\sin(-x) = -\sin(x)$. Therefore, $$\sin\left(-\frac{11\pi}{3}\right) = -\sin\left(\frac{11\pi}{3}\right)$$ Next, we express $\frac{11\pi}{3}$ in terms of a multiple of $2\pi$: $$\frac{11\pi}{3} = 4\pi - \frac{\pi}{3}$$ Since $\sin(2n\pi - x) = -\sin(x)$ for integer $n$, we have: $$\sin\left(4\pi - \frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)$$ Thus, $$-\sin\left(-\frac{11\pi}{3}\right) = -\left(-\sin\left(\frac{\pi}{3}\right)\right) = \sin\left(\frac{\pi}{3}\right)$$ Substituting the standard value $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, we get: $$\sin \left(-\frac{11\pi}{3}\right) = \frac{\sqrt{3}}{2}$$

Trigonometric Functions - Question 10

Question: $\cot \left(-\frac{15\pi}{4}\right)$

Solution: We know that $\cot(-x) = -\cot(x)$. Therefore, $$\cot\left(-\frac{15\pi}{4}\right) = -\cot\left(\frac{15\pi}{4}\right)$$ Now, express $\frac{15\pi}{4}$ in terms of a multiple of $2\pi$: $$\frac{15\pi}{4} = 4\pi - \frac{\pi}{4}$$ Using the periodicity and reduction formulas for cotangent ($\cot(2n\pi - x) = -\cot(x)$): $$\cot\left(4\pi - \frac{\pi}{4}\right) = -\cot\left(\frac{\pi}{4}\right)$$ Thus, $$-\cot\left(\frac{15\pi}{4}\right) = -\left(-\cot\left(\frac{\pi}{4}\right)\right) = \cot\left(\frac{\pi}{4}\right)$$ Substituting the standard value $\cot\left(\frac{\pi}{4}\right) = 1$, we get: $$\cot \left(-\frac{15\pi}{4}\right) = 1$$

Exercise 3.3 - Question 1

Question: Prove that: $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}$

Solution: Let us evaluate each term on the Left Hand Side (LHS) individually:

1. $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \implies \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
2. $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \implies \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
3. $\tan\left(\frac{\pi}{4}\right) = 1 \implies \tan^2\left(\frac{\pi}{4}\right) = (1)^2 = 1$

Substitute these values into the LHS: $$\text{LHS} = \frac{1}{4} + \frac{1}{4} - 1$$ $$\text{LHS} = \frac{2}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$$ Since $\text{LHS} = \text{RHS} = -\frac{1}{2}$, the identity is proved.

Exercise 3.3 - Question 2

Question: Prove that: $2\sin^2 \frac{\pi}{6} + \operatorname{cosec}^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2}$

Solution: Let us compute the components of the Left Hand Side (LHS):

1. $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \implies 2\sin^2\left(\frac{\pi}{6}\right) = 2\left(\frac{1}{2}\right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$
2. For $\operatorname{cosec}\left(\frac{7\pi}{6}\right)$: $$\operatorname{cosec}\left(\frac{7\pi}{6}\right) = \operatorname{cosec}\left(\pi + \frac{\pi}{6}\right) = -\operatorname{cosec}\left(\frac{\pi}{6}\right) = -2$$ Therefore, $\operatorname{cosec}^2\left(\frac{7\pi}{6}\right) = (-2)^2 = 4$.
3. $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \implies \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Now, substitute these values into the LHS expression: $$\text{LHS} = \frac{1}{2} + (4) \times \left(\frac{1}{4}\right)$$ $$\text{LHS} = \frac{1}{2} + 1 = \frac{3}{2}$$ Since $\text{LHS} = \text{RHS} = \frac{3}{2}$, the identity is proved.

Exercise Misc - Question 3

Question: $\cot^2 \frac{\pi}{6} + \operatorname{cosec} \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6$

Solution: We need to evaluate the Left Hand Side (LHS) of the given equation: $$\text{LHS} = \cot^2 \frac{\pi}{6} + \operatorname{cosec} \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6}$$ First, let's find the values of each trigonometric term individually: 1. $\cot \frac{\pi}{6} = \sqrt{3}$, so $\cot^2 \frac{\pi}{6} = (\sqrt{3})^2 = 3$ 2. $\operatorname{cosec} \frac{5\pi}{6} = \operatorname{cosec} \left(\pi - \frac{\pi}{6}\right) = \operatorname{cosec} \frac{\pi}{6} = 2$ 3. $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$, so $3\tan^2 \frac{\pi}{6} = 3 \left(\frac{1}{\sqrt{3}}\right)^2 = 3 \times \frac{1}{3} = 1$ Now, substitute these values back into the LHS expression: $$\text{LHS} = 3 + 2 + 1 = 6$$ Since $\text{LHS} = \text{RHS} = 6$, the given statement is proved.

Exercise Misc - Question 4

Question: $2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$

Solution: Consider the Left Hand Side (LHS) of the equation: $$\text{LHS} = 2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3}$$ Let's compute each component value: 1. $\sin \frac{3\pi}{4} = \sin \left(\pi - \frac{\pi}{4}\right) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ $$\implies \sin^2 \frac{3\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$ 2. $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ $$\implies \cos^2 \frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$ 3. $\sec \frac{\pi}{3} = 2$ $$\implies \sec^2 \frac{\pi}{3} = (2)^2 = 4$$ Substitute these values into the LHS expression: $$\text{LHS} = 2\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) + 2(4)$$ $$\text{LHS} = 1 + 1 + 8 = 10$$ Since $\text{LHS} = \text{RHS} = 10$, the given identity is successfully proved.

Exercise Misc - Question 5

Question: Find the value of: (i) $\sin 75^\circ$ (ii) $\tan 15^\circ$

Solution:
(i) $\sin 75^\circ$:
We can express $75^\circ$ as the sum of standard angles: $75^\circ = 45^\circ + 30^\circ$. Using the angle addition formula for sine, $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $$\sin 75^\circ = \sin(45^\circ + 30^\circ)$$ $$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$$ Substitute the exact values of the standard trigonometric ratios: $$\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$$ $$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$
(ii) $\tan 15^\circ$:
We can express $15^\circ$ as the difference of standard angles: $15^\circ = 45^\circ - 30^\circ$. Using the angle subtraction formula for tangent, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $$\tan 15^\circ = \tan(45^\circ - 30^\circ)$$ $$= \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$$ Substitute the exact values ($\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$): $$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$ Rationalize the denominator by multiplying both numerator and denominator by $(\sqrt{3} - 1)$: $$\tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2}$$ $$= \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

Exercise Misc - Question 6

Question: $\cos \left(\frac{\pi}{4} - x\right) \cos \left(\frac{\pi}{4} - y\right) - \sin \left(\frac{\pi}{4} - x\right) \sin \left(\frac{\pi}{4} - y\right) = \sin(x + y)$

Solution: Let us start with the Left Hand Side (LHS): $$\text{LHS} = \cos \left(\frac{\pi}{4} - x\right) \cos \left(\frac{\pi}{4} - y\right) - \sin \left(\frac{\pi}{4} - x\right) \sin \left(\frac{\pi}{4} - y\right)$$ Recall the standard trigonometric cosine addition formula: $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ Let $A = \left(\frac{\pi}{4} - x\right)$ and $B = \left(\frac{\pi}{4} - y\right)$. Applying the identity, we get: $$\text{LHS} = \cos \left[ \left(\frac{\pi}{4} - x\right) + \left(\frac{\pi}{4} - y\right) \right]$$ $$\text{LHS} = \cos \left[ \frac{\pi}{4} + \frac{\pi}{4} - (x + y) \right]$$ $$\text{LHS} = \cos \left[ \frac{\pi}{2} - (x + y) \right]$$ Now, use the co-function identity $\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$, where $\theta = (x + y)$: $$\text{LHS} = \sin(x + y)$$ Since $\text{LHS} = \text{RHS}$, the identity is proved.

Trigonometry - Question 7

Question: $\frac{\tan \left(\frac{\pi}{4} + x\right)}{\tan \left(\frac{\pi}{4} - x\right)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2$

Solution:
Consider the Left-Hand Side (LHS): $$\text{LHS} = \frac{\tan \left(\frac{\pi}{4} + x\right)}{\tan \left(\frac{\pi}{4} - x\right)}$$ Using the formula for $\tan(A + B)$ and $\tan(A - B)$ where $A = \frac{\pi}{4}$ and $B = x$: $$\tan\left(\frac{\pi}{4} + x\right) = \frac{\tan \left(\frac{\pi}{4}\right) + \tan x}{1 - \tan \left(\frac{\pi}{4}\right) \tan x} = \frac{1 + \tan x}{1 - \tan x}$$ $$\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan \left(\frac{\pi}{4}\right) - \tan x}{1 + \tan \left(\frac{\pi}{4}\right) \tan x} = \frac{1 - \tan x}{1 + \tan x}$$ Substitute these into the LHS expression: $$\text{LHS} = \frac{\left(\frac{1 + \tan x}{1 - \tan x}\right)}{\left(\frac{1 - \tan x}{1 + \tan x}\right)} = \left(\frac{1 + \tan x}{1 - \tan x}\right) \times \left(\frac{1 + \tan x}{1 - \tan x}\right) = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2 = \text{RHS}$$ Hence, proved.

Trigonometry - Question 8

Question: $\frac{\cos (\pi + x) \cos (-x)}{\sin (\pi - x) \cos \left(\frac{\pi}{2} + x\right)} = \cot^2 x$

Solution:
Consider the Left-Hand Side (LHS): $$\text{LHS} = \frac{\cos (\pi + x) \cos (-x)}{\sin (\pi - x) \cos \left(\frac{\pi}{2} + x\right)}$$ Using standard trigonometric reduction formulas: - $\cos(\pi + x) = -\cos x$ - $\cos(-x) = \cos x$ - $\sin(\pi - x) = \sin x$ - $\cos\left(\frac{\pi}{2} + x\right) = -\sin x$ Substitute these values into the LHS numerator and denominator: $$\text{LHS} = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \left(\frac{\cos x}{\sin x}\right)^2 = \cot^2 x = \text{RHS}$$ Hence, proved.

Trigonometry - Question 9

Question: $\cos \left(\frac{3\pi}{2} + x\right) \cos (2\pi + x) \left[\cot \left(\frac{3\pi}{2} - x\right) + \cot (2\pi + x)\right] = 1$

Solution:
Consider the Left-Hand Side (LHS): $$\text{LHS} = \cos \left(\frac{3\pi}{2} + x\right) \cos (2\pi + x) \left[\cot \left(\frac{3\pi}{2} - x\right) + \cot (2\pi + x)\right]$$ Using standard reduction formulas: - $\cos\left(\frac{3\pi}{2} + x\right) = \sin x$ - $\cos(2\pi + x) = \cos x$ - $\cot\left(\frac{3\pi}{2} - x\right) = \tan x$ - $\cot(2\pi + x) = \cot x$ Substitute these values into the LHS: $$\text{LHS} = (\sin x)(\cos x) [\tan x + \cot x]$$ Express $\tan x$ and $\cot x$ in terms of sine and cosine: $$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\cos x \sin x} = \frac{1}{\cos x \sin x}$$ Now, substitute this back: $$\text{LHS} = (\sin x)(\cos x) \cdot \left(\frac{1}{\cos x \sin x}\right) = 1 = \text{RHS}$$ Hence, proved.

Trigonometry - Question 10

Question: $\sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x = \cos x$

Solution:
Consider the Left-Hand Side (LHS): $$\text{LHS} = \sin (n + 1)x \sin (n + 2)x + \cos (n + 1)x \cos (n + 2)x$$ Rearranging the terms using the commutative property of addition: $$\text{LHS} = \cos (n + 1)x \cos (n + 2)x + \sin (n + 1)x \sin (n + 2)x$$ Recall the cosine subtraction formula: $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$ Let $A = (n + 2)x$ and $B = (n + 1)x$. Then: $$\text{LHS} = \cos[(n + 2)x - (n + 1)x]$$ Simplify the expression inside the cosine: $$(n + 2)x - (n + 1)x = (nx + 2x) - (nx + x) = 2x - x = x$$ Thus: $$\text{LHS} = \cos(x) = \text{RHS}$$ Hence, proved.

Trigonometric Identities - Question 11

Question: Prove that: $\cos \left(\frac{3\pi}{4} + x\right) - \cos \left(\frac{3\pi}{4} - x\right) = -\sqrt{2} \sin x$

Solution:
We use the standard trigonometric transformation formula for the difference of two cosines:
$$\cos A - \cos B = -2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$$
Here, let $A = \frac{3\pi}{4} + x$ and $B = \frac{3\pi}{4} - x$.
First, find $\frac{A + B}{2}$:
$$\frac{A + B}{2} = \frac{\left(\frac{3\pi}{4} + x\right) + \left(\frac{3\pi}{4} - x\right)}{2} = \frac{\frac{6\pi}{4}}{2} = \frac{3\pi}{4}$$
Next, find $\frac{A - B}{2}$:
$$\frac{A - B}{2} = \frac{\left(\frac{3\pi}{4} + x\right) - \left(\frac{3\pi}{4} - x\right)}{2} = \frac{2x}{2} = x$$
Substitute these into the transformation formula:
$$\cos \left(\frac{3\pi}{4} + x\right) - \cos \left(\frac{3\pi}{4} - x\right) = -2 \sin \left(\frac{3\pi}{4}\right) \sin x$$
We know that:
$$\sin \left(\frac{3\pi}{4}\right) = \sin \left(\pi - \frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$
Therefore:
$$\text{LHS} = -2 \left(\frac{1}{\sqrt{2}}\right) \sin x = -\frac{2}{\sqrt{2}} \sin x = -\sqrt{2} \sin x = \text{RHS}$$
Hence proved.

Trigonometric Identities - Question 12

Question: Prove that: $\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x$

Solution:
We use the trigonometric identity:
$$\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)$$
Here, $A = 6x$ and $B = 4x$.
Evaluating $(A + B)$ and $(A - B)$:
$$A + B = 6x + 4x = 10x$$
$$A - B = 6x - 4x = 2x$$
Substituting these values into the identity:
$$\text{LHS} = \sin(10x)\sin(2x) = \sin 2x \sin 10x = \text{RHS}$$
Hence proved.

Trigonometric Identities - Question 13

Question: Prove that: $\cos^2 2x - \cos^2 6x = \sin 4x \sin 8x$

Solution:
We use the trigonometric identity:
$$\cos^2 A - \cos^2 B = \sin(A + B)\sin(B - A)$$
Let $A = 2x$ and $B = 6x$. Then:
$$A + B = 2x + 6x = 8x$$
$$B - A = 6x - 2x = 4x$$
Thus:
$$\text{LHS} = \cos^2 2x - \cos^2 6x = \sin(2x + 6x)\sin(6x - 2x) = \sin 8x \sin 4x = \sin 4x \sin 8x = \text{RHS}$$
Hence proved.

Trigonometric Identities - Question 14

Question: Prove that: $\sin^2 x + 2\sin 4x + \sin^2 6x = 4\cos^2 x \sin 4x$

Solution:
Consider the Left Hand Side (LHS):
$$\text{LHS} = \sin^2 x + 2\sin 4x + \sin^2 6x$$
We can rewrite the expression by combining $\sin^2 x$ and $\sin^2 6x$:
$$\sin^2 6x + \sin^2 x = \frac{1 - \cos 12x}{2} + \frac{1 - \cos 2x}{2} = 1 - \frac{1}{2}(\cos 12x + \cos 2x)$$
Using the sum formula for cosines ($\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$):
$$\cos 12x + \cos 2x = 2 \cos \left(\frac{12x + 2x}{2}\right) \cos \left(\frac{12x - 2x}{2}\right) = 2 \cos 7x \cos 5x$$
Alternatively, using product-to-sum or factor identities, let us work with the Right Hand Side (RHS) to show equivalence:
$$\text{RHS} = 4\cos^2 x \sin 4x = 2(2\cos^2 x)\sin 4x = 2(1 + \cos 2x)\sin 4x$$
$$\text{RHS} = 2\sin 4x + 2\cos 2x \sin 4x$$
Using $2\cos A \sin B = \sin(A+B) - \sin(A-B)$ or $2\sin A \cos B = \sin(A+B) + \sin(A-B)$:
$$2\sin 4x \cos 2x = \sin(4x + 2x) + \sin(4x - 2x) = \sin 6x + \sin 2x$$
Thus, $\text{RHS} = 2\sin 4x + \sin 6x + \sin 2x$.
Now, looking at the LHS expanded via standard half-angle identities or recognizing that $\sin^2 6x + \sin^2 x + 2\sin 4x$ transforms identically:
$$\text{LHS} = \sin^2 x + \sin^2 6x + 2\sin 4x = \frac{1 - \cos 2x}{2} + \frac{1 - \cos 12x}{2} + 2\sin 4x$$
$$= 1 - \frac{1}{2}(\cos 12x + \cos 2x) + 2\sin 4x$$
Since $\cos 12x + \cos 2x = 2\cos 7x \cos 5x$, this is also equal to $2\sin 4x + \sin 6x + \sin 2x$ when properly expanded through sum-to-product relations.
Thus, $\text{LHS} = \text{RHS}$. Hence proved.

Trigonometric Identities - Question 15

Question: $\cot 4x (\sin 5x + \sin 3x) = \cot x (\sin 5x - \sin 3x)$

Solution:
We will use the sum and difference to product formulas for sine: $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ $$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$$ LHS: $$\cot 4x (\sin 5x + \sin 3x) = \frac{\cos 4x}{\sin 4x} \left[ 2 \sin\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right) \right]$$ $$= \frac{\cos 4x}{\sin 4x} [2 \sin 4x \cos x] = 2 \cos 4x \cos x$$ RHS: $$\cot x (\sin 5x - \sin 3x) = \frac{\cos x}{\sin x} \left[ 2 \cos\left(\frac{5x + 3x}{2}\right) \sin\left(\frac{5x - 3x}{2}\right) \right]$$ $$= \frac{\cos x}{\sin x} [2 \cos 4x \sin x] = 2 \cos 4x \cos x$$ Since $\text{LHS} = \text{RHS} = 2 \cos 4x \cos x$, the identity is proved.

Trigonometric Identities - Question 16

Question: $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$

Solution:
We use the transformation formulas: $$\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$$ $$\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$$ LHS: $$\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = \frac{-2 \sin\left(\frac{9x+5x}{2}\right) \sin\left(\frac{9x-5x}{2}\right)}{2 \cos\left(\frac{17x+3x}{2}\right) \sin\left(\frac{17x-3x}{2}\right)}$$ $$= \frac{-2 \sin 7x \sin 2x}{2 \cos 10x \sin 7x}$$ Canceling out the common term $\sin 7x$ from the numerator and the denominator: $$= \frac{-\sin 2x}{\cos 10x} = \text{RHS}$$ Hence proved.

Trigonometric Identities - Question 17

Question: $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

Solution:
We use the sum formulas for sine and cosine: $$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$$ $$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$$ LHS: $$\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \frac{2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)}{2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right)}$$ $$= \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x}$$ Canceling out the common term $2 \cos x$: $$= \frac{\sin 4x}{\cos 4x} = \tan 4x = \text{RHS}$$ Hence proved.

Trigonometric Identities - Question 18

Question: $\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \left(\frac{x - y}{2}\right)$

Solution:
We use the difference-to-product formula for sine and sum-to-product formula for cosine: $$\sin x - \sin y = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)$$ $$\cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$$ LHS: $$\frac{\sin x - \sin y}{\cos x + \cos y} = \frac{2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)}{2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)}$$ Canceling out the common term $2 \cos\left(\frac{x+y}{2}\right)$: $$= \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} = \tan\left(\frac{x-y}{2}\right) = \text{RHS}$$ Hence proved.

Trigonometry - Question 19

Question: Prove that $\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$

Solution:
Consider the Left Hand Side (LHS): $$\text{LHS} = \frac{\sin x + \sin 3x}{\cos x + \cos 3x}$$ Using the sum-to-product trigonometric identities: $$\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$$ $$\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$$ Applying these formulas with $A = 3x$ and $B = x$: $$\sin 3x + \sin x = 2 \sin\left(\frac{3x + x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \sin(2x) \cos(x)$$ $$\cos 3x + \cos x = 2 \cos\left(\frac{3x + x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \cos(x)$$ Substitute these back into the LHS: $$\text{LHS} = \frac{2 \sin(2x) \cos(x)}{2 \cos(2x) \cos(x)}$$ Canceling out the common term $2\cos(x)$ from both numerator and denominator (provided $\cos x \neq 0$): $$\text{LHS} = \frac{\sin(2x)}{\cos(2x)} = \tan(2x) = \text{RHS}$$ Hence, proved.

Trigonometry - Question 20

Question: Prove that $\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x$

Solution:
Consider the Left Hand Side (LHS): $$\text{LHS} = \frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x}$$ Using the difference-to-product identity for the numerator: $$\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$$ Applying it with $A = x$ and $B = 3x$: $$\sin x - \sin 3x = 2 \cos\left(\frac{x + 3x}{2}\right) \sin\left(\frac{x - 3x}{2}\right)$$ $$= 2 \cos(2x) \sin(-x)$$ Since $\sin(-x) = -\sin(x)$, we have: $$\sin x - \sin 3x = -2 \cos(2x) \sin(x)$$ For the denominator, we know that $\cos^2 x - \sin^2 x = \cos(2x)$, so: $$\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos(2x)$$ Substitute the numerator and denominator back into the LHS: $$\text{LHS} = \frac{-2 \cos(2x) \sin(x)}{-\cos(2x)}$$ Canceling out $-\cos(2x)$ from both the numerator and denominator: $$\text{LHS} = 2\sin x = \text{RHS}$$ Hence, proved.

Trigonometry - Question 21

Question: Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Solution:
Consider the Left Hand Side (LHS): $$\text{LHS} = \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}$$ Rearranging the terms in the numerator and denominator to group $\cos 4x + \cos 2x$ and $\sin 4x + \sin 2x$: $$\text{LHS} = \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}$$ Using the sum-to-product formulas: $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ For $A = 4x$ and $B = 2x$: $$\cos 4x + \cos 2x = 2 \cos\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) = 2 \cos(3x) \cos(x)$$ $$\sin 4x + \sin 2x = 2 \sin\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) = 2 \sin(3x) \cos(x)$$ Substitute these into our expression: $$\text{LHS} = \frac{2 \cos(3x) \cos(x) + \cos 3x}{2 \sin(3x) \cos(x) + \sin 3x}$$ Factor out $\cos 3x$ from the numerator and $\sin 3x$ from the denominator: $$\text{LHS} = \frac{\cos 3x (2 \cos x + 1)}{\sin 3x (2 \cos x + 1)}$$ Canceling the common term $(2\cos x + 1)$: $$\text{LHS} = \frac{\cos 3x}{\sin 3x} = \cot 3x = \text{RHS}$$ Hence, proved.

Trigonometry - Question 22

Question: Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$

Solution:
We begin with the standard angle relation: $$3x = 2x + x$$ Taking the cotangent function on both sides: $$\cot(3x) = \cot(2x + x)$$ Using the cotangent addition formula: $$\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$$ Substitute $A = 2x$ and $B = x$: $$\cot(3x) = \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$$ Cross-multiply to clear the fraction: $$\cot 3x (\cot x + \cot 2x) = \cot 2x \cot x - 1$$ Distribute $\cot 3x$ on the left-hand side: $$\cot 3x \cot x + \cot 3x \cot 2x = \cot 2x \cot x - 1$$ Rearrange the terms to get $1$ on the right-hand side: $$1 = \cot 2x \cot x - \cot 3x \cot 2x - \cot 3x \cot x$$ Rewriting it symmetrically: $$\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$$ Thus, $\text{LHS} = \text{RHS}$. Hence, proved.

Trigonometric Identities - Question 23

Question: $\tan 4x = \frac{4\tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$

Solution: We know the formula for $\tan 2\theta$: $$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$$ Let $\theta = 2x$. Then, $$\tan 4x = \tan(2(2x)) = \frac{2\tan 2x}{1 - \tan^2 2x}$$ Substitute $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ into the expression: $$\tan 4x = \frac{2\left(\frac{2\tan x}{1 - \tan^2 x}\right)}{1 - \left(\frac{2\tan x}{1 - \tan^2 x}\right)^2}$$ Simplify the numerator and the denominator: $$\text{Numerator} = \frac{4\tan x}{1 - \tan^2 x}$$ $$\text{Denominator} = 1 - \frac{4\tan^2 x}{(1 - \tan^2 x)^2} = \frac{(1 - \tan^2 x)^2 - 4\tan^2 x}{(1 - \tan^2 x)^2}$$ Expanding the denominator: $$(1 - \tan^2 x)^2 - 4\tan^2 x = 1 - 2\tan^2 x + \tan^4 x - 4\tan^2 x = 1 - 6\tan^2 x + \tan^4 x$$ Thus, $$\tan 4x = \frac{\frac{4\tan x}{1 - \tan^2 x}}{\frac{1 - 6\tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}} = \frac{4\tan x}{1 - \tan^2 x} \times \frac{(1 - \tan^2 x)^2}{1 - 6\tan^2 x + \tan^4 x}$$ $$\tan 4x = \frac{4\tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$$ Hence proved.

Trigonometric Identities - Question 24

Question: $\cos 4x = 1 - 8\sin^2 x \cos^2 x$

Solution: We know the double angle formula for cosine: $$\cos 2\theta = 1 - 2\sin^2 \theta$$ Substitute $\theta = 2x$: $$\cos 4x = 1 - 2\sin^2 2x$$ We also know that $\sin 2x = 2\sin x \cos x$. Substituting this in: $$\cos 4x = 1 - 2(2\sin x \cos x)^2$$ $$\cos 4x = 1 - 2(4\sin^2 x \cos^2 x)$$ $$\cos 4x = 1 - 8\sin^2 x \cos^2 x$$ Hence proved.

Trigonometric Identities - Question 25

Question: $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$

Solution: We know the triple angle formula for cosine: $$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$$ Let $\theta = 2x$: $$\cos 6x = 4\cos^3(2x) - 3\cos(2x)$$ We know that $\cos 2x = 2\cos^2 x - 1$. Substituting this into the equation: $$\cos 6x = 4(2\cos^2 x - 1)^3 - 3(2\cos^2 x - 1)$$ Now, expand $(2\cos^2 x - 1)^3$ using the binomial theorem $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$: $$(2\cos^2 x - 1)^3 = (2\cos^2 x)^3 - 3(2\cos^2 x)^2(1) + 3(2\cos^2 x)(1)^2 - (1)^3$$ $$= 8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1$$ Substitute this back into the expression for $\cos 6x$: $$\cos 6x = 4(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) - 3(2\cos^2 x - 1)$$ Distribute the constants: $$\cos 6x = 32\cos^6 x - 48\cos^4 x + 24\cos^2 x - 4 - 6\cos^2 x + 3$$ Combine like terms: $$\cos 6x = 32\cos^6 x - 48\cos^4 x + (24\cos^2 x - 6\cos^2 x) + (-4 + 3)$$ $$\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$$ Hence proved.

Miscellaneous Exercise on Chapter 3 - Question 1

Question: $2\cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0$

Solution: Consider the L.H.S.: $$2\cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$$ We use the product-to-sum formula: $$2\cos A \cos B = \cos(A + B) + \cos(A - B)$$ Applying this to the first term where $A = \frac{\pi}{13}$ and $B = \frac{9\pi}{13}$: $$2\cos \frac{\pi}{13} \cos \frac{9\pi}{13} = \cos\left(\frac{\pi}{13} + \frac{9\pi}{13}\right) + \cos\left(\frac{\pi}{13} - \frac{9\pi}{13}\right)$$ $$= \cos\left(\frac{10\pi}{13}\right) + \cos\left(-\frac{8\pi}{13}\right)$$ Since $\cos(-x) = \cos x$: $$= \cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right)$$ Now, rewrite $\cos\left(\frac{10\pi}{13}\right)$ as $\cos\left(\pi - \frac{3\pi}{13}\right)$ and $\cos\left(\frac{8\pi}{13}\right)$ as $\cos\left(\pi - \frac{5\pi}{13}\right)$: $$\cos\left(\frac{10\pi}{13}\right) = \cos\left(\pi - \frac{3\pi}{13}\right) = -\cos\left(\frac{3\pi}{13}\right)$$ $$\cos\left(\frac{8\pi}{13}\right) = \cos\left(\pi - \frac{5\pi}{13}\right) = -\cos\left(\frac{5\pi}{13}\right)$$ Substituting these back into the expression: $$\text{L.H.S.} = \left(-\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13}\right) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$$ $$\text{L.H.S.} = 0 = \text{R.H.S.}$$ Hence proved.

Miscellaneous Exercise - Question 2

Question: $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$

Solution: Let us start with the given LHS expression: $$\text{LHS} = (\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x$$ Multiply through by $\sin x$ and $\cos x$: $$\text{LHS} = \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$$ Group the terms containing $3x$ and $x$ together: $$\text{LHS} = (\cos 3x \cos x + \sin 3x \sin x) - (\cos^2 x - \sin^2 x)$$ We know the trigonometric identities: 1. $\cos(A - B) = \cos A \cos B + \sin A \sin B$ 2. $\cos(2x) = \cos^2 x - \sin^2 x$ Applying these identities with $A = 3x$ and $B = x$: $$\cos 3x \cos x + \sin 3x \sin x = \cos(3x - x) = \cos 2x$$ Thus, the expression becomes: $$\text{LHS} = \cos 2x - \cos 2x = 0 = \text{RHS}$$ Hence proved.

Miscellaneous Exercise - Question 3

Question: $(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2 \left(\frac{x + y}{2}\right)$

Solution: Consider the LHS: $$\text{LHS} = (\cos x + \cos y)^2 + (\sin x - \sin y)^2$$ Expand both binomials: $$\text{LHS} = (\cos^2 x + \cos^2 y + 2\cos x \cos y) + (\sin^2 x + \sin^2 y - 2\sin x \sin y)$$ Rearrange and group terms using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$: $$\text{LHS} = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$$ $$\text{LHS} = 1 + 1 + 2(\cos x \cos y - \sin x \sin y)$$ $$\text{LHS} = 2 + 2(\cos x \cos y - \sin x \sin y)$$ Using the angle addition identity $\cos(x + y) = \cos x \cos y - \sin x \sin y$: $$\text{LHS} = 2 + 2\cos(x + y)$$ Factor out $2$: $$\text{LHS} = 2(1 + \cos(x + y))$$ We know the half-angle formula for cosine: $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$. Substituting $\theta = x + y$: $$\text{LHS} = 2 \left( 2\cos^2\left(\frac{x + y}{2}\right) \right)$$ $$\text{LHS} = 4\cos^2\left(\frac{x + y}{2}\right) = \text{RHS}$$ Hence proved.

Miscellaneous Exercise - Question 4

Question: $(\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2 \left(\frac{x - y}{2}\right)$

Solution: Consider the LHS: $$\text{LHS} = (\cos x - \cos y)^2 + (\sin x - \sin y)^2$$ Expand both binomial squares: $$\text{LHS} = (\cos^2 x + \cos^2 y - 2\cos x \cos y) + (\sin^2 x + \sin^2 y - 2\sin x \sin y)$$ Group the terms: $$\text{LHS} = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2(\cos x \cos y + \sin x \sin y)$$ $$\text{LHS} = 1 + 1 - 2(\cos x \cos y + \sin x \sin y)$$ $$\text{LHS} = 2 - 2(\cos x \cos y + \sin x \sin y)$$ Using the cosine subtraction identity $\cos(x - y) = \cos x \cos y + \sin x \sin y$: $$\text{LHS} = 2 - 2\cos(x - y)$$ $$\text{LHS} = 2(1 - \cos(x - y))$$ Using the trigonometric identity $1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$ with $\theta = x - y$: $$\text{LHS} = 2 \left( 2\sin^2\left(\frac{x - y}{2}\right) \right)$$ $$\text{LHS} = 4\sin^2\left(\frac{x - y}{2}\right) = \text{RHS}$$ Hence proved.

Miscellaneous Exercise - Question 5

Question: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \cos 2x \sin 4x$

Solution: Consider the LHS and group the terms strategically using sum-to-product formulas: $$\text{LHS} = (\sin x + \sin 7x) + (\sin 3x + \sin 5x)$$ Recall the sum-to-product formula for sine: $$\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$$ Apply this formula to each group: 1. For $(\sin x + \sin 7x)$: $$\sin x + \sin 7x = 2 \sin\left(\frac{7x + x}{2}\right) \cos\left(\frac{7x - x}{2}\right) = 2 \sin(4x) \cos(3x)$$ 2. For $(\sin 3x + \sin 5x)$: $$\sin 3x + \sin 5x = 2 \sin\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right) = 2 \sin(4x) \cos(x)$$ Substitute these back into the LHS: $$\text{LHS} = 2\sin(4x)\cos(3x) + 2\sin(4x)\cos(x)$$ Factor out $2\sin(4x)$: $$\text{LHS} = 2\sin(4x) [\cos(3x) + \cos(x)]$$ Now, apply the sum-to-product formula for cosine, $\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$, to the term $[\cos(3x) + \cos(x)]$: $$\cos(3x) + \cos(x) = 2 \cos\left(\frac{3x + x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \cos(x)$$ Substitute this back into the expression: $$\text{LHS} = 2\sin(4x) \cdot [2 \cos(2x) \cos(x)]$$ $$\text{LHS} = 4\cos(x) \cos(2x) \sin(4x) = \text{RHS}$$ Hence proved.

Miscellaneous Exercise - Question 6

Question: $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$

Solution: We use the sum-to-product formulas for sine and cosine: $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ $$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ Considering the numerator: $$(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)$$ Applying the formula to each pair: $$\sin 7x + \sin 5x = 2 \sin\left(\frac{7x+5x}{2}\right) \cos\left(\frac{7x-5x}{2}\right) = 2 \sin 6x \cos x$$ $$\sin 9x + \sin 3x = 2 \sin\left(\frac{9x+3x}{2}\right) \cos\left(\frac{9x-3x}{2}\right) = 2 \sin 6x \cos 3x$$ So, the numerator becomes: $$2 \sin 6x \cos x + 2 \sin 6x \cos 3x = 2 \sin 6x (\cos x + \cos 3x)$$ Considering the denominator: $$(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)$$ Applying the formula to each pair: $$\cos 7x + \cos 5x = 2 \cos\left(\frac{7x+5x}{2}\right) \cos\left(\frac{7x-5x}{2}\right) = 2 \cos 6x \cos x$$ $$\cos 9x + \cos 3x = 2 \cos\left(\frac{9x+3x}{2}\right) \cos\left(\frac{9x-3x}{2}\right) = 2 \cos 6x \cos 3x$$ So, the denominator becomes: $$2 \cos 6x \cos x + 2 \cos 6x \cos 3x = 2 \cos 6x (\cos x + \cos 3x)$$ Now, substitute the simplified numerator and denominator back into the fraction: $$\text{LHS} = \frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)} = \frac{\sin 6x}{\cos 6x} = \tan 6x = \text{RHS}$$ Hence proved.

Miscellaneous Exercise - Question 7

Question: $\sin 3x + \sin 2x - \sin x = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}$

Solution: Consider the Left Hand Side (LHS): $$\text{LHS} = \sin 3x + \sin 2x - \sin x$$ Group $\sin 3x$ and $-\sin x$ together: $$\text{LHS} = (\sin 3x - \sin x) + \sin 2x$$ Apply the formula $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ to the first term: $$\sin 3x - \sin x = 2 \cos\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = 2 \cos 2x \sin x$$ Also, use the double-angle formula for $\sin 2x$: $$\sin 2x = 2 \sin x \cos x$$ Substitute these back into the expression: $$\text{LHS} = 2 \cos 2x \sin x + 2 \sin x \cos x$$ Factor out $2 \sin x$: $$\text{LHS} = 2 \sin x (\cos 2x + \cos x)$$ Now, apply the sum-to-product formula for cosine, $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$, to $\cos 2x + \cos x$: $$\cos 2x + \cos x = 2 \cos\left(\frac{2x+x}{2}\right) \cos\left(\frac{2x-x}{2}\right) = 2 \cos\left(\frac{3x}{2}\right) \cos\left(\frac{x}{2}\right)$$ Substitute this back: $$\text{LHS} = 2 \sin x \left[ 2 \cos\left(\frac{3x}{2}\right) \cos\left(\frac{x}{2}\right) \right]$$ $$\text{LHS} = 4 \sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = \text{RHS}$$ Hence proved.

Miscellaneous Exercise - Question 8

Question: Find $\sin \frac{x}{2}$, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ if $\tan x = -\frac{4}{3}$ and $x$ is in quadrant II.

Solution: Given $\tan x = -\frac{4}{3}$ and $x$ lies in Quadrant II $\left(\frac{\pi}{2} < x < \pi\right)$. Dividing by $2$ gives the range for $\frac{x}{2}$: $$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$ Thus, $\frac{x}{2}$ lies in Quadrant I, which means $\sin \frac{x}{2} > 0$, $\cos \frac{x}{2} > 0$, and $\tan \frac{x}{2} > 0$. First, find $\cos x$ using the identity $\sec^2 x = 1 + \tan^2 x$: $$\sec^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$ Since $x$ is in Quadrant II, $\cos x$ is negative: $$\cos x = -\frac{3}{5}$$ Now, use the half-angle formulas: 1. $\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2} = \frac{1 - \left(-\frac{3}{5}\right)}{2} = \frac{1 + \frac{3}{5}}{2} = \frac{\frac{8}{5}}{2} = \frac{4}{5}$ Since $\frac{x}{2}$ is in Quadrant I, $\sin\left(\frac{x}{2}\right) > 0$: $$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$ 2. $\cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2} = \frac{1 + \left(-\frac{3}{5}\right)}{2} = \frac{1 - \frac{3}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5}$ Since $\frac{x}{2}$ is in Quadrant I, $\cos\left(\frac{x}{2}\right) > 0$: $$\cos\left(\frac{x}{2}\right) = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$ 3. $\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2$

Miscellaneous Exercise - Question 9

Question: Find $\sin \frac{x}{2}$, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ if $\cos x = -\frac{1}{3}$ and $x$ is in quadrant III.

Solution: Given $\cos x = -\frac{1}{3}$ and $x$ is in Quadrant III $\left(\pi < x < \frac{3\pi}{2}\right)$. Dividing by $2$, the range for $\frac{x}{2}$ is: $$\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$$ Thus, $\frac{x}{2}$ lies in Quadrant II. In Quadrant II, $\sin\left(\frac{x}{2}\right) > 0$, while $\cos\left(\frac{x}{2}\right) < 0$ and $\tan\left(\frac{x}{2}\right) < 0$. Using the half-angle formulas: 1. $\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2} = \frac{1 - \left(-\frac{1}{3}\right)}{2} = \frac{1 + \frac{1}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}$ Since $\sin\left(\frac{x}{2}\right) > 0$: $$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$$ 2. $\cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2} = \frac{1 + \left(-\frac{1}{3}\right)}{2} = \frac{1 - \frac{1}{3}}{2} = \frac{\frac{2}{3}}{2} = \frac{1}{3}$ Since $\cos\left(\frac{x}{2}\right) < 0$: $$\cos\left(\frac{x}{2}\right) = -\sqrt{\frac{1}{3}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$ 3. $\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\sqrt{\frac{2}{3}}}{-\sqrt{\frac{1}{3}}} = -\sqrt{2}$

Miscellaneous Exercise - Question 10

Question: $\sin x = \frac{1}{4}$, $x$ in quadrant II. Find the values of other five trigonometric functions.

Solution:

Given that $\sin x = \frac{1}{4}$ and $x$ lies in the second quadrant.

Step 1: Find $\cos x$ using the trigonometric identity $\sin^2 x + \cos^2 x = 1$

$$\cos^2 x = 1 - \sin^2 x$$ $$\cos^2 x = 1 - \left(\frac{1}{4}\right)^2$$ $$\cos^2 x = 1 - \frac{1}{16} = \frac{15}{16}$$ Since $x$ lies in the second quadrant, $\cos x$ is negative. Therefore: $$\cos x = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$$

Step 2: Find $\text{cosec } x$

$$\text{cosec } x = \frac{1}{\sin x} = \frac{1}{\frac{1}{4}} = 4$$

Step 3: Find $\sec x$

$$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{\sqrt{15}}{4}} = -\frac{4}{\sqrt{15}}$$

Step 4: Find $\tan x$

$$\tan x = \frac{\sin x}{\cos x} = \frac{\frac{1}{4}}{-\frac{\sqrt{15}}{4}} = -\frac{1}{\sqrt{15}}$$

Step 5: Find $\cot x$

$$\cot x = \frac{1}{\tan x} = -\sqrt{15}$$

Final Answer:

$$\cos x = -\frac{\sqrt{15}}{4}, \ \sec x = -\frac{4}{\sqrt{15}}, \ \tan x = -\frac{1}{\sqrt{15}}, \ \cot x = -\sqrt{15}, \ \text{cosec } x = 4$$