Question
Prove that $\frac{\tan A}{\left(1+\tan ^2 A\right)^2}+\frac{\cot A}{\left(1+\cot ^2 A\right)^2}=\sin A \cdot \cos A$
✓
Answer
$\text { LHS }=\frac{\tan A}{\left(1+\tan ^2 A\right)^2}+\frac{\cot A}{\left(1+\cot ^2 A\right)^2}$
$=\frac{\tan A}{\left(\sec ^2 A\right)^2}+\frac{\cot A}{(\operatorname{cosec} A)^2}$
$=\frac{\sin A}{\cos A} \times \cos ^2 A \times \cos ^2 A+\frac{\cos A}{\sin A} \times \sin ^2 A \times \sin ^2 A$
$= \sin A.\cos^3A + \sin^3A.\cos A$
$= \sin A \cos A (\cos^2A + \sin^2A)$
$= \sin A. \cos A x 1$
$= \sin A. \cos A$
$= RHS$
Hence proved.
$=\frac{\tan A}{\left(\sec ^2 A\right)^2}+\frac{\cot A}{(\operatorname{cosec} A)^2}$
$=\frac{\sin A}{\cos A} \times \cos ^2 A \times \cos ^2 A+\frac{\cos A}{\sin A} \times \sin ^2 A \times \sin ^2 A$
$= \sin A.\cos^3A + \sin^3A.\cos A$
$= \sin A \cos A (\cos^2A + \sin^2A)$
$= \sin A. \cos A x 1$
$= \sin A. \cos A$
$= RHS$
Hence proved.
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