$ \sqrt{\sin ^4 x+4 \cos ^2 x}-\sqrt{\cos ^4 x+4 \sin ^2 x}= $

$ \sqrt{\sin ^4 x+4 \cos ^2 x}-\sqrt{\cos ^4 x+4 \sin ^2 x}= $
  1. $1-\cos 2 x$
  2. $\tan 2 x$
  3. $\sin 2 x$
  4. $\cos 2 x$

Solution

$\begin{aligned} & \text { We have } \sqrt{\sin ^4+4 \cos ^2 x}-\sqrt{\cos ^4 x+4 \sin ^2 x} \\ & \Rightarrow \sqrt{\left(2-\sin ^2(x)\right)^2}-\sqrt{\left(2-\cos ^2 x\right)^2} \\ & \Rightarrow 2-\sin ^2(x)-\left(2-\cos ^2(x)\right) \\ & =\cos ^2(x)-\sin ^2(x) \\ & =\cos 2 x\end{aligned}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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