# Prove the following identities:$\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)$

To do:

We have to prove that $\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)$.

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

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

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

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

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

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

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

Hence proved.

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Updated on: 10-Oct-2022

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