Prove the following trigonometric identities.

We need to prove $\sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1$

Solving the L.H.S, we get

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

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

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

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

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

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

$=1+\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta$    (using $1+\tan ^{2} \theta=\sec ^{2} \theta$ )

Hence proved.