Question

$\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }=\sec \theta cosec\theta -2\sin \theta \cos \theta$.

Answer 1

$\frac{{\tan }^3\theta }{\left(1+{\tan }^2\theta \right)}+\frac{{\cot }^3\theta }{\left(1+{\cot }^2\theta \right)}=\sec \theta .\cos ec\theta -2\sin \theta .\cos \theta \Rightarrow \frac{{\tan }^3\theta }{{\sec }^2\theta }+\frac{{\cot }^3\theta }{\cos e{c}^2\theta }\Rightarrow \frac{{\sin }^3\theta }{{\cos }^3\theta .{\sec }^2\theta }+\frac{{\cos }^3\theta }{{\sin }^3\theta .\cos e{c}^2\theta }\Rightarrow \frac{{\sin }^3\theta }{\cos \theta }+\frac{{\cos }^3\theta }{\sin \theta }\Rightarrow \frac{{\sin }^4\theta +{\cos }^4\theta }{\sin \theta .\cos \theta }=\frac{{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2}{\sin \theta .\cos \theta }\Rightarrow \frac{{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2{\sin }^2\theta .{\cos }^2\theta }{\sin \theta .\cos \theta }=\frac{1-2{\sin }^2\theta .{\cos }^2\theta }{\sin \theta .\cos \theta }\therefore \Rightarrow \left[\frac{1}{\sin \theta .\cos }-\frac{2{\sin }^2\theta .{\cos }^2\theta }{\sin \theta .\cos \theta }\right]=\sec \theta .\cos ec\theta -2\sin \theta .\cos \theta$