JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 15)
If $\sin x + \sin^2 x = 1$, $x \in \left(0, \frac{\pi}{2}\right)$, then
$(\cos^{12} x + \tan^{12} x) + 3(\cos^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to:
3
4
2
1
Explanation
$$\begin{aligned}
&\begin{aligned}
& \sin x+\sin ^2 x=1 \\
& \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x
\end{aligned}\\
&\therefore \text { Given expression }\\
&\begin{aligned}
& =2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x \\
& =2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right] \\
& =2 \sin ^3 x\left[(\sin x+1)^3\right] \\
& =2\left[\sin ^2 x+\sin x\right]^3 \\
& =2
\end{aligned}
\end{aligned}$$
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