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

To do:

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

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,

$\frac{\tan A}{\left(1+\tan ^{2} A\right)^{2}}+\frac{\cot A}{\left(1+\cot ^{2} A\right)^{2}}=\frac{\tan A}{(\sec ^{2} A)^{2}}+\frac{\cot A}{(\operatorname{cosec} ^{2} A)^{2}}$

$=\frac{\sin A}{\cos A} \times \frac{\cos ^{4} A}{1}+\frac{\cos A}{\sin A} \times \frac{\sin ^{4} A}{1}$

$=\sin A \cos ^{3} A+\sin ^{3} A \cos A$

$=\sin A \cos A\left(\cos ^{2} A+\sin ^{2} A\right)$

$=\sin A \cos A(1)$

$=\sin A \cos A$

Hence proved.

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Updated on: 10-Oct-2022

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