# If $\cos \theta+\cos ^{2} \theta=1$, prove that$\sin ^{12} \theta+3 \sin ^{10} \theta+3 \sin ^{8} \theta+\sin ^{6} \theta+2 \sin ^{4} \theta+2 \sin ^{2} \theta-2=1$

Given:

$\cos \theta+\cos ^{2} \theta=1$

To do:

We have to prove that $\sin ^{12} \theta+3 \sin ^{10} \theta+3 \sin ^{8} \theta+\sin ^{6} \theta+2 \sin ^{4} \theta+2 \sin ^{2} \theta-2=1$.

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$\cos \theta+\cos ^{2} \theta=1$

$\Rightarrow \cos \theta=1-\cos ^{2} \theta$

$\Rightarrow \cos \theta=\sin ^{2} \theta$

$\Rightarrow \cos^2 \theta=\sin ^{4} \theta$

This implies,

$\sin ^{12} \theta+3 \sin ^{10} \theta+3 \sin ^{8} \theta+\sin ^{6} \theta +2 \sin ^{4} \theta+2 \sin ^{2} \theta-2 =(\sin ^{12} \theta+3 \sin ^{10} \theta+3 \sin ^{8} \theta+\sin ^{6} \theta)+2(\sin ^{4} \theta+\sin ^{2} \theta-1)$

$=(\sin ^{4} \theta+\sin ^{2} \theta)^{3}+2(\sin ^{4} \theta+\sin ^{2} \theta-1)$       [Since $(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$]

$=(\cos ^{2} \theta+\sin ^{2} \theta)^{3}+2(\cos ^{2} \theta+\sin ^{2} \theta-1)$

$=(1)^{3}+2(1-1)$

$=1+2 \times 0$

$=1+0$

$=1$

Hence proved.

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Updated on: 10-Oct-2022

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