If $\cos heta, \sin heta$ and $\cot heta$ are in geometric progression, then $\sin ^9 heta+\sin ^6 heta+3 \sin ^5 heta+\sin ^3 heta+\sin ^2 heta=$

Question

If $\cos 	heta, \sin 	heta$ and $\cot 	heta$ are in geometric progression, then $\sin ^9 	heta+\sin ^6 	heta+3 \sin ^5 	heta+\sin ^3 	heta+\sin ^2 	heta=$

Aicharya · Accepted Answer

$\because \cos 	heta, \sin 	heta, \cot 	heta$ are in geometric progression. $ \begin{aligned} & 	herefore \frac{\sin 	heta}{\cos 	heta}=\frac{\cot 	heta}{\sin 	heta} \ & \Rightarrow \frac{\sin 	heta}{\cos 	heta}=\frac{\cos 	heta}{\sin ^2 	heta} \Rightarrow \sin ^3 	heta=\cos ^2 	heta.....(i) \end{aligned} $ $ \begin{aligned} & 	ext { Now, } \sin ^9 	heta+\sin ^6 	heta+3 \sin ^5 	heta+\sin ^3 	heta+\sin ^2 	heta \ & =\cos ^6 	heta+\sin ^6 	heta+3 \sin ^5 	heta+\cos ^2 	heta+\sin ^2 	heta \quad 	ext { [from (i)] } \ & =\left(\cos ^2 	heta+\sin ^2 	hetaight)\left(\cos ^4 	heta+\sin ^4 	heta-\cos ^2 	heta \sin ^2 	hetaight)+3 \ & \sin ^5 	heta+1 \ & =\left(\cos ^2 	heta+\sin ^2 	hetaight)^2-3 \sin ^2 	heta \cos ^2 	heta+3 \sin ^5 	heta+1 \ & =1^2-3 \sin ^5 	heta+3 \sin ^5 	heta+1=2 \end{aligned} $

If $\cos \theta, \sin \theta$ and $\cot \theta$ are in geometric progression, then $\sin ^9 \theta+\sin ^6…

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