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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.