(A) $\dfrac{5}{\sqrt{13}}$

(B) $\dfrac{6}{\sqrt{26}}$

(C) $\dfrac{4}{\sqrt{26}}$

(D) $\dfrac{1}{\sqrt{13}}$

Correct Answer: $\dfrac{1}{\sqrt{13}}$

Explanation

Let

$$ A=\cos\frac{13x}{2}, \quad B=\sin\frac{13x}{2} $$

Given expression becomes

$$ \sin7x(A+B)+\cos7x(A-B) $$

$$ =A(\sin7x+\cos7x)+B(\sin7x-\cos7x) $$

Using identities

$$ \sin7x+\cos7x=\sqrt{2}\sin\left(7x+\frac{\pi}{4}\right), $$

$$ \sin7x-\cos7x=\sqrt{2}\sin\left(7x-\frac{\pi}{4}\right) $$

So the expression becomes

$$ \sqrt{2}\left[ \cos\frac{13x}{2}\sin\left(7x+\frac{\pi}{4}\right) +\sin\frac{13x}{2}\sin\left(7x-\frac{\pi}{4}\right) \right] $$

Since

$$ \sin\left(7x-\frac{\pi}{4}\right) =\sin\left(7x+\frac{\pi}{4}-\frac{\pi}{2}\right) =-\cos\left(7x+\frac{\pi}{4}\right), $$

the bracket becomes

$$ \sin\left(7x+\frac{\pi}{4}\right)\cos\frac{13x}{2} -\cos\left(7x+\frac{\pi}{4}\right)\sin\frac{13x}{2} $$

$$ =\sin\left(7x+\frac{\pi}{4}-\frac{13x}{2}\right) $$

Hence the value is

$$ \sqrt{2}\sin\left(\frac{x}{2}+\frac{\pi}{4}\right) $$

Given $\cot x=\frac{5}{12}$ and $x\in(\pi,\frac{3\pi}{2})$,

$$ \sin x=-\frac{12}{13}, \quad \cos x=-\frac{5}{13} $$

Half-angle values:

$$ \sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\frac{3}{\sqrt{13}}, \quad \cos\frac{x}{2}=-\sqrt{\frac{1+\cos x}{2}}=-\frac{2}{\sqrt{13}} $$

Now

$$ \sin\left(\frac{x}{2}+\frac{\pi}{4}\right) =\sin\frac{x}{2}\cos\frac{\pi}{4}+\cos\frac{x}{2}\sin\frac{\pi}{4} $$

$$ =\frac{1}{\sqrt{2}}\left(\frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}\right) =\frac{1}{\sqrt{26}} $$

Therefore,

$$ \text{Required value} =\sqrt{2}\cdot\frac{1}{\sqrt{26}} =\frac{1}{\sqrt{13}} $$

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