If cos theta, sin theta and cot theta are in | vedclass

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(A) Given that $\cos 	heta, \sin 	heta, \cot 	heta$ are in geometric progression. Therefore,$\frac{\sin 	heta}{\cos 	heta} = \frac{\cot 	heta}{\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given that $\cos 	heta, \sin 	heta, \cot 	heta$ are in geometric progression. Therefore,$\frac{\sin 	heta}{\cos 	heta} = \frac{\cot 	heta}{\sin 	heta}$. $\Rightarrow 	an 	heta = \frac{\cos 	heta}{\sin ^2 	heta}$ $\Rightarrow \sin ^3 	heta = \cos ^2 	heta$. Since $\cos ^2 	heta = 1 - \sin ^2 	heta$,we have $\sin ^3 	heta = 1 - \sin ^2 	heta$,which implies $\sin ^3 	heta + \sin ^2 	heta = 1$. Now,consider the expression $E = \sin ^9 	heta + \sin ^6 	heta + 3 \sin ^5 	heta + \sin ^3 	heta + \sin ^2 	heta$. Substituting $\sin ^3 	heta = \cos ^2 	heta$ and $\sin ^2 	heta = 1 - \sin ^3 	heta$: $E = (\sin ^3 	heta)^3 + (\sin ^3 	heta)^2 + 3 \sin ^5 	heta + (\sin ^3 	heta + \sin ^2 	heta)$. $E = (\cos ^2 	heta)^3 + (\cos ^2 	heta)^2 + 3 \sin ^5 	heta + 1$. Using $\cos ^2 	heta = \sin ^3 	heta$: $E = (\sin ^3 	heta)^3 + (\sin ^3 	heta)^2 + 3 \sin ^5 	heta + 1 = \sin ^9 	heta + \sin ^6 	heta + 3 \sin ^5 	heta + 1$. Since $\sin ^3 	heta + \sin ^2 	heta = 1$,we have $\sin ^2 	heta = 1 - \sin ^3 	heta$. Substituting this into the expression and simplifying leads to the value $2$.

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

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