Question
Prove that: $\frac{\tan A }{\left(1+\tan ^2 A \right)^2}+\frac{\cot A }{\left(1+\cot ^2 A \right)^2}=\sin A \cos A$
✓
Answer
Taking RHS
$\begin{array}{l}\frac{\tan A }{\left(1+\tan ^2 A \right)^2}+\frac{\cot A }{\left(1+\cot ^2 A \right)^2} \\
=\frac{\tan A }{\left(\sec ^2 A \right)^2}+\frac{\cot A }{\left(\operatorname{cosec}^2 A \right)^2} \\
=\frac{\tan A }{\sec ^4 A }+\frac{\cot A }{\operatorname{cosec}^4 A } \\
=\frac{\sin A}{\cos A} \cdot \cos ^4 A+\frac{\cos A}{\sin A} \cdot \sin ^4 A \\
=\sin A \cos ^3 A+\cos A \sin ^3 A \\
=\sin A \cos A\left(\cos ^2 A+\sin ^2 A\right)\left[A s, \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\sin A \cos A \\
= RHS \\
\end{array}$
Proved!
$\begin{array}{l}\frac{\tan A }{\left(1+\tan ^2 A \right)^2}+\frac{\cot A }{\left(1+\cot ^2 A \right)^2} \\
=\frac{\tan A }{\left(\sec ^2 A \right)^2}+\frac{\cot A }{\left(\operatorname{cosec}^2 A \right)^2} \\
=\frac{\tan A }{\sec ^4 A }+\frac{\cot A }{\operatorname{cosec}^4 A } \\
=\frac{\sin A}{\cos A} \cdot \cos ^4 A+\frac{\cos A}{\sin A} \cdot \sin ^4 A \\
=\sin A \cos ^3 A+\cos A \sin ^3 A \\
=\sin A \cos A\left(\cos ^2 A+\sin ^2 A\right)\left[A s, \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\sin A \cos A \\
= RHS \\
\end{array}$
Proved!
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