If $\sin x+\sin ^2 x=1, x \in\left(0, \frac{\pi}{2} ight)$, then $\left(\cos ^{12} x+ an ^{12} x ight)+3\left(\cos ^{10} x+ an ^{10} x+ ight.$ $\left.\cos ^8 x+ an ^8 x ight)+\left(\cos ^6 x+ an ^6 x ight)$ is equal to :

Question

If $\sin x+\sin ^2 x=1, x \in\left(0, \frac{\pi}{2}ight)$, then $\left(\cos ^{12} x+	an ^{12} xight)+3\left(\cos ^{10} x+	an ^{10} x+ight.$ $\left.\cos ^8 x+	an ^8 xight)+\left(\cos ^6 x+	an ^6 xight)$ is equal to :

Aicharya · Accepted Answer

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

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

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