JEE Main 2025 — Trigonometric Ratios & Identities Question with Solution

$\begin{aligned} & \sin x+\sin ^2 x=1 \\ & \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x \end{aligned}$ $\therefore$ Given expression $\begin{aligned} & =2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x \\ & =2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right] \\ & =2 \sin ^3 x\left[(\sin x+1)^3\right] \\ & =2\left[\sin ^2 x+\sin x\right]^3 \\ & =2 \end{aligned}$