# Prove that:$\left(1+\tan ^{2} A\right)+\left(1+\frac{1}{\tan ^{2} A}\right)=\frac{1}{\sin ^{2} A-\sin ^{4} A}$

To do:

We have to prove that $\left(1+\tan ^{2} A\right)+\left(1+\frac{1}{\tan ^{2} A}\right)=\frac{1}{\sin ^{2} A-\sin ^{4} A}$.

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,

Let us consider LHS,

$\left(1+\tan ^{2} \mathrm{~A}\right)\left(1+\frac{1}{\tan ^{2} \mathrm{~A}}\right)=\left(1+\tan ^{2} \mathrm{~A}\right)+\left(1+\cot ^{2} \mathrm{~A}\right)$

$=\sec ^{2} \mathrm{~A} \operatorname{cosec}^{2} \mathrm{~A}$

Let us consider RHS,

$=\frac{1}{\sin ^{2} \mathrm{~A}-\sin ^{4} \mathrm{~A}}=\frac{1}{\sin ^{2} \mathrm{~A}-\sin ^{4} \mathrm{~A}}$

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

$=\frac{1}{\sin ^{2} A \cos ^{2} A}$

$=\operatorname{cosec}^{2} A \sec ^{2} A$

$=\sec ^{2} A \operatorname{cosec}^{2} A$

Here,

LHS $=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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