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3.Trigonometrical Ratios, Functions and Identities
hard
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
A$4$
B$3$
C$2$
D$1$
(JEE MAIN-2025)
Solution
$\sin x+\sin ^2 x=1$
$\Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x$
$\therefore$ Given expression
$=2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x$
$=2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right]$
$=2 \sin ^3 x\left[(\sin x+1)^3\right]$
$=2\left[\sin ^2 x+\sin x\right]^3$
$=2$
$\Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x$
$\therefore$ Given expression
$=2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x$
$=2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right]$
$=2 \sin ^3 x\left[(\sin x+1)^3\right]$
$=2\left[\sin ^2 x+\sin x\right]^3$
$=2$
Standard 11
Mathematics
