Prove the following trigonometric identities:$ \operatorname{cosec}^{6} \theta=\cot ^{6} \theta+3 \cot ^{2} \theta \operatorname{cosec}^{2} \theta+1 $

To do:

We have to prove that \( \operatorname{cosec}^{6} \theta=\cot ^{6} \theta+3 \cot ^{2} \theta \operatorname{cosec}^{2} \theta+1 \).

Solution:

We know that,

$\operatorname{cosec} ^{2} \theta-\cot^2 \theta=1$.......(i)

$(a+b)^3=a^3+b^3+3ab(a+b)$.........(ii)

Therefore,

$\operatorname{cosec}^{6} \theta=(\operatorname{cosec} ^2 \theta)^3$

$=(1+\cot^2 \theta)^3$                [From (i)]

$=1^3+(\cot^2 \theta)^3+3(1)(\cot^2 \theta)(1+\cot^2 \theta)$        [From (ii)]          

$=1+\cot^6 \theta+3\cot^2 \theta(1+\cot^2 \theta)$

$=1+\cot^6 \theta+3\cot^2 \theta\operatorname{cosec}^2 \theta$         [From (i)]

$=\cot ^{6} \theta+3 \cot ^{2} \theta \operatorname{cosec}^{2} \theta+1$

Hence proved.    

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Updated on: 10-Oct-2022

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