Question
Prove the trigonometric identity: $\frac{\tan \theta}{\left(1+\tan ^2 \theta\right)^2}+\frac{\cot \theta}{\left(1+\cot ^2 \theta\right)^2}=\sin \theta \cos \theta$
✓
Answer
$\frac{\tan \theta}{\left(1+\tan ^2 \theta\right)^2}+\frac{\cot \theta}{\left(1+\cot ^2 \theta\right)^2}$
$=\frac{\tan \theta}{\left(\sec ^2 \theta\right)^2}+\frac{\cot \theta}{\left(\operatorname{cosec}^2 \theta\right)^2}$
$=\frac{\sin \theta}{\cos \theta} \times \frac{1}{\sec ^4 \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{1}{\operatorname{cosec} 4}$
$=\frac{\sin \theta}{\cos \theta} \times \cos ^4 \theta+\frac{\cos \theta}{\sin \theta} \times \sin ^4 \theta$
$=\sin \theta \cos ^3 \theta+\cos \theta \sin ^3 \theta$
$=\sin \theta \cos \theta\left(\cos ^2 \theta+\sin ^2 \theta\right)$
$=\sin \theta \cos \theta=\text { RHS }$
Hence proved.
$=\frac{\tan \theta}{\left(\sec ^2 \theta\right)^2}+\frac{\cot \theta}{\left(\operatorname{cosec}^2 \theta\right)^2}$
$=\frac{\sin \theta}{\cos \theta} \times \frac{1}{\sec ^4 \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{1}{\operatorname{cosec} 4}$
$=\frac{\sin \theta}{\cos \theta} \times \cos ^4 \theta+\frac{\cos \theta}{\sin \theta} \times \sin ^4 \theta$
$=\sin \theta \cos ^3 \theta+\cos \theta \sin ^3 \theta$
$=\sin \theta \cos \theta\left(\cos ^2 \theta+\sin ^2 \theta\right)$
$=\sin \theta \cos \theta=\text { RHS }$
Hence proved.
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