#### Question

Prove that:

\[\frac{\tan A}{\left( 1 + \tan^2 A \right)^2} + \frac{\cot A}{\left( 1 + \cot^2 A \right)^2} = \sin A \cos A\]

#### Solution

\[\frac{\tan A}{\left( 1 + \tan^2 A \right)^2} + \frac{\cot A}{\left( 1 + \cot^2 A \right)^2}\]

\[ = \frac{\tan A}{\left( \sec^2 A \right)^2} + \frac{\cot A}{\left( {cosec}^2 A \right)^2} \left( 1 + \tan^2 \theta = \sec^2 \theta\text{ and }1 + \cot^2 \theta = {cosec}^2 \theta \right)\]

\[ = \frac{\sin A}{\cos A } \times \cos^4 A + \frac{\cos A}{\sin A} \times \sin^4 A \left( \cos\theta = \frac{1}{\sec\theta}\text{ and }\sin\theta = \frac{1}{cosec\theta} \right)\]

\[ = \sin A \cos^3 A + \cos A \sin^3 A\]

\[= \sin A\cos A\left( \cos^2 A + \sin^2 A \right)\]

\[ = \sin A\cos A \left( \cos^2 \theta + \sin^2 \theta = 1 \right)\]

Is there an error in this question or solution?

Solution Prave That: Tan a ( 1 + Tan 2 a ) 2 + Cot a ( 1 + Cot 2 a ) 2 = Sin a Cos a Concept: Application of Trigonometry.