Question

$\frac{\tan \theta }{(1+{\tan }^2\theta {)}^2}+\frac{\cot \theta }{(1+{\cot }^2\theta {)}^2}=\sin \theta \cos \theta$.

Answer 1

$\frac{\tan \theta }{\left(1+{\tan }^2\theta \right)}+\frac{\cot \theta }{{\left(1+{\cot }^2\theta \right)}^2}=\sin \theta .\cos \theta \left[\begin{array}{c}\because 1+{\tan }^2\theta ={\sec }^2\theta \\ 1+{\cot }^2\theta =\cos e{c}^2\theta \end{array}\right]\Rightarrow \frac{\tan \theta }{{\left({\sec }^2\theta \right)}^2}+\frac{\cot \theta }{{\left(\cos e{c}^2\theta \right)}^2}\Rightarrow \frac{\sin \theta }{\cos \theta .\sec \theta .{\sec }^3\theta }+\frac{\cos \theta }{\sin \theta .\cos ec\theta .\cos e{c}^3\theta }\Rightarrow \frac{\sin \theta }{{\sec }^3\theta }+\frac{\cos \theta }{\cos e{c}^3\theta }\left[\begin{array}{c}\because \cos \theta .\sec \theta =1\\ \cos ec\theta .\sin \theta =1\end{array}\right]\Rightarrow \sin \theta .{\cos }^3\theta +\cos \theta .{\sin }^3\theta \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\Rightarrow \sin \theta .\cos \theta .\left[{\cos }^2\theta ={\sin }^3\theta \right]\therefore \sin \theta .\cos \theta \times 1=\sin \theta .\cos \theta$