Q70 / 180 · COMEDK 2015 · Mathematics

$\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$ is equal to

  • A
    $-2 \sin x$
  • B
    $\frac{2}{\sin x}$
  • C
    $\frac{1}{\sin x}$
  • Correct answer
    D
    $2 \sin x$

Solution

$\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$ $$ \begin{aligned} &=\frac{2 \sin \left(\frac{x-3 x}{2}\right) \cos \left(\frac{x+3 x}{2}\right)}{-\left(\cos ^{2} x-\sin ^{2} x\right)} \\ &=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}=\frac{-2 \sin x \cos 2 x}{-\cos 2 x}=2 \sin x \end{aligned} $$

Hence, the correct answer is option: 'D'.

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