Prove the following trigonometric identities.

In the given question, we need to prove

$\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)$

Now, using the identity $(a+b)^{2}=a^{2}+b^{2}+2 a b$ in L.H.S, we get

$\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=\left(\tan ^{2} \theta+\frac{1}{\cos ^{2} \theta}+2 \frac{\tan \theta}{\cos \theta}\right)+\left(\tan ^{2} \theta+\frac{1}{\cos ^{2} \theta}-2 \frac{\tan \theta}{\cos \theta}\right)$

$=\tan ^{2} \theta+\frac{1}{\cos ^{2} \theta}+2 \frac{\tan \theta}{\cos \theta}+\tan ^{2} \theta+\frac{1}{\cos ^{2} \theta}-2 \frac{\tan \theta}{\cos \theta}$

$=2 \tan ^{2} \theta+\frac{2}{\cos ^{2} \theta}$

Further using $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$2 \tan ^{2} \theta+\frac{2}{\cos ^{2} \theta}=\frac{2 \sin ^{2} \theta}{\cos ^{2} \theta}+\frac{2}{\cos ^{2} \theta}$

$=\frac{2 \sin ^{2} \theta+2}{\cos ^{2} \theta}$

$=\frac{2\left(\sin ^{2} \theta+1\right)}{\cos ^{2} \theta}$

Also, from the identity $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

$\frac{2\left(\sin ^{2} \theta+1\right)}{\cos ^{2} \theta}=\frac{2\left(\sin ^{2} \theta+1\right)}{1-\sin ^{2} \theta}$

$=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)$

Hence proved.