Trigonometric Ratio and Identites — JEE Main practice

1. Start with the given expression:

   $$\operatorname{cosec}10^\circ-\sqrt3\sec10^\circ =\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}.$$

2. Take the LCM:

   $$=\frac{\cos10^\circ-\sqrt3\sin10^\circ} {\sin10^\circ\cos10^\circ}.$$

3. Multiply the numerator and denominator appropriately:

   $$=4\left[ \frac{\frac12\cos10^\circ-\frac{\sqrt3}{2}\sin10^\circ} {2\sin10^\circ\cos10^\circ} \right].$$

4. Use the identities

   $$\sin(A-B)=\sin A\cos B-\cos A\sin B,$$

   and

   $$2\sin10^\circ\cos10^\circ=\sin20^\circ.$$

   Since

   $$\frac12=\sin30^\circ,\qquad \frac{\sqrt3}{2}=\cos30^\circ,$$

   the numerator becomes

   $$\sin(30^\circ-10^\circ)=\sin20^\circ.$$

5. Therefore,

   $$4\left[ \frac{\sin20^\circ}{\sin20^\circ} \right] =4.$$

6. Hence, the required value is

   $$\boxed{4}.$$

Therefore, Option C is correct.

1.1. Let

$$\begin{array}{rcl} E&=&\sin\left(\frac{15\theta}{2}\right) \left(\cos8\theta+\sin8\theta\right)\\[6pt] &&+\cos\left(\frac{15\theta}{2}\right) \left(\cos8\theta-\sin8\theta\right). \end{array}$$

2. Expand the expression:

$$\begin{array}{rcl} E &=&\sin\frac{15\theta}{2}\cos8\theta +\sin\frac{15\theta}{2}\sin8\theta\\[6pt] &&+\cos\frac{15\theta}{2}\cos8\theta -\cos\frac{15\theta}{2}\sin8\theta. \end{array}$$

3. Group the terms and use the identities

$$\cos A\cos B+\sin A\sin B=\cos(A-B),$$

and

$$\sin A\cos B-\cos A\sin B=\sin(A-B).$$

Thus,

$$\begin{array}{rcl} E &=&\cos\left(8\theta-\frac{15\theta}{2}\right) +\sin\left(\frac{15\theta}{2}-8\theta\right)\\[6pt] &=&\cos\frac{\theta}{2} -\sin\frac{\theta}{2}. \end{array}$$

4. Given

$$\cot\theta=-\frac{1}{2\sqrt2}, \qquad \frac{\pi}{2}<\theta<\pi,$$

the source derives

$$\cos\theta=-\frac13.$$

5. Apply the half-angle identities:

$$\cos\theta=1-2\sin^2\frac{\theta}{2},$$

so

$$\begin{array}{rcl} -\dfrac13 &=&1-2\sin^2\dfrac{\theta}{2}\\[6pt] \Rightarrow\quad \sin\dfrac{\theta}{2} &=&\sqrt{\dfrac23}. \end{array}$$

6. Also,

$$\cos\theta=2\cos^2\frac{\theta}{2}-1,$$

giving

$$\begin{array}{rcl} -\dfrac13 &=&2\cos^2\dfrac{\theta}{2}-1\\[6pt] \Rightarrow\quad \cos\dfrac{\theta}{2} &=&\sqrt{\dfrac13}. \end{array}$$

7. Therefore,

$$\begin{array}{rcl} E &=&\cos\dfrac{\theta}{2} -\sin\dfrac{\theta}{2}\\[6pt] &=&\sqrt{\dfrac13}-\sqrt{\dfrac23}\\[6pt] &=&\dfrac{1-\sqrt2}{\sqrt3}. \end{array}$$

8. Hence, the required value is

$$\boxed{\frac{1-\sqrt2}{\sqrt3}}.$$

Therefore, Option B is correct.

$$\begin{array}{l} E=\sin7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right)\\[6pt] \qquad+\cos7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right). \end{array}$$

2. Expand the expression:

$$\begin{array}{rcl} E &=&\sin7x\cos\frac{13x}{2} +\sin7x\sin\frac{13x}{2}\\[6pt] &&+\cos7x\cos\frac{13x}{2} -\cos7x\sin\frac{13x}{2}. \end{array}$$

3. Rearrange the terms and apply the identities

$$\sin A\cos B-\cos A\sin B=\sin(A-B),$$

and

$$\cos A\cos B+\sin A\sin B=\cos(A-B).$$

Therefore,

$$\begin{array}{rcl} E &=&\sin\left(7x-\frac{13x}{2}\right) +\cos\left(7x-\frac{13x}{2}\right)\\[6pt] &=&\sin\frac{x}{2}+\cos\frac{x}{2}. \end{array}$$

4. Let

$$L=\sin\frac{x}{2}+\cos\frac{x}{2}.$$

Since

$$x\in\left(\pi,\frac{3\pi}{2}\right),$$

we have

$$\frac{x}{2}\in\left(\frac{\pi}{2},\frac{3\pi}{4}\right),$$

so $L$ is positive.

5. Square both sides:

$$\begin{array}{rcl} L^2 &=&\sin^2\frac{x}{2} +\cos^2\frac{x}{2} +2\sin\frac{x}{2}\cos\frac{x}{2}\\[6pt] &=&1+\sin x. \end{array}$$

6. Given

$$\cot x=\frac{5}{12},$$

and $x$ lies in the third quadrant, $\sin x<0$.

Hence,

$$\begin{array}{rcl} \sin x &=&-\dfrac{1}{\sqrt{1+\cot^2x}}\\[6pt] &=&-\dfrac{1}{\sqrt{1+\frac{25}{144}}}\\[6pt] &=&-\dfrac{12}{13}. \end{array}$$

7. Therefore,

$$\begin{array}{rcl} L^2 &=&1-\dfrac{12}{13}\\[6pt] &=&\dfrac{1}{13}. \end{array}$$

Since $L>0$,

$$L=\frac{1}{\sqrt{13}}.$$

8. Thus,

$$E=\frac{1}{\sqrt{13}}.$$

Hence, Option A is correct.