Question

$\frac{\tan \theta }{1+{\tan }^2\theta }+\frac{\cot \theta }{(1+{\cot }^2\theta {)}^2}$ is equal to

• A. $2\sin \theta \cdot \cos \theta$
• B. $\text{cosec}\theta \cdot \sec \theta$
• C. $\sin \theta \cdot \cos \theta$
• D. $2\text{cosec}\theta \cdot \sec \theta$

Answer 1

Answer: (C)

$\sin \theta \cdot \cos \theta$

$\frac{\tan \theta }{(1+{\tan }^2\theta {)}^2}+\frac{\cot \theta }{(1+{\cot }^2\theta {)}^2}=\frac{\tan \theta }{{\sec }^4\theta }+\frac{\cot \theta }{{\text{cosec}}^4\theta }$

$=\frac{\text{cosec}\theta \sec \theta (\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta })}{{\sec }^4\theta {\text{cosec}}^4\theta }=\frac{{\sec }^2\theta {\text{cosec}}^2\theta }{{\sec }^3\theta {\text{cosec}}^3\theta }=\sin \theta \cos \theta$.