If $\sin x+\sin ^2 x=1, x \in\left(0, \frac{\pi}{2}\right)$, then $\left(\cos ^{12} x+\tan ^{12} x\right)+3\left(\cos ^{10} x+\tan ^{10} x+\cos ^8 x+\tan ^8 x\right)+\left(\cos ^6 x+\tan ^6 x\right)$ is equal to: [JEE Main 2025]

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YESPYQ · Accepted Answer

Correct answer: a) 2. $\sin x+\sin ^2 x=1$ $\Rightarrow \sin x=\cos ^2 x$ and $\tan x=\cos x$ Given expression becomes $=2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x$ $=2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right]$ $=2 \sin ^3 x\left[(\sin x+1)^3\right]$ $=2\left[\sin ^2 x+\sin x\right]^3$ $=2$

If \(\sin x+\sin ^2 x=1, x \in\left(0, \frac{\pi}{2}\right)\), then \(\left(\cos ^{12} x+\tan ^{12} x\right)+3\left(\cos…

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