If \sin\theta+\sin^{2}\theta=1 , then the value of \cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1 is equal to

  • Option 1)

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  • Option 2)

    1

  • Option 3)

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  • Option 4)

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Answers (1)

 

Trigonometric Identities -

\sin ^{2}\Theta + \cos ^{2}\Theta = 1

1 + \tan ^{2}\Theta = \sec ^{2}\Theta

1 + \cot ^{2}\Theta = cosec ^{2}\Theta

- wherein

They are true for all real values of \Theta

 

 \sin \theta +\sin ^{2}\theta = 1

\cos ^{12}\theta +3\cos ^{10}\theta +3\cos ^{8}\theta +\cos ^{6}\theta -1 =

\Rightarrow 1-\sin ^{2}\theta =\sin \theta

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

\therefore \cos ^{12}\theta = \sin ^{6}\theta

\cos ^{10}\theta = \sin ^{5}\theta

\cos ^{8}\theta = \sin ^{4}\theta

\cos ^{6}\theta =\sin ^{3}\theta

\Rightarrow \sin ^{6}\theta +3\sin ^{5}\theta +3\sin ^{4}\theta +\sin ^{3}\theta -1

\Rightarrow \left ( \sin ^{2} \Theta \right )^{3} + \left ( \sin \theta \right )^{3} + 3 \sin \theta \cdot \sin ^{2}\theta (\sin \theta +\sin ^{2}\theta )-1

= \left ( 1 \right )^{3}-1\Rightarrow 1-1=0


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