# Prove the following trigonometric identities

Question:

Prove the following trigonometric identities

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

Solution:

We need to prove $\operatorname{cosec}^{6} \theta=\cot ^{6} \theta+3 \cot ^{2} \theta \operatorname{cosec}^{2} \theta+1$

Solving the L.H.S, we get

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

$=\left(1+\cot ^{2} \theta\right)^{3} \quad\left(1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\right)$

Further using the identity $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$, we get

$\left(1+\cot ^{2} \theta\right)^{3}=1+\cot ^{6} \theta+3(1)^{2}\left(\cot ^{2} \theta\right)+3(1)\left(\cot ^{2} \theta\right)^{2}$

$=1+\cot ^{6} \theta+3 \cot ^{2} \theta+3 \cot ^{4} \theta$

$=1+\cot ^{6} \theta+3 \cot ^{2} \theta\left(1+\cot ^{2} \theta\right)$

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

Hence proved.