Question

Find $\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }-\sec \theta \csc \theta +2\sin \theta \cos \theta$

Answer 1

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

$\because 1+{\tan }^{2}x={\sec }^{2}x$

$1+{\cot }^{2}x={\cos ec}^{2}x$

$\sec x=\frac{1}{\cos x}$

$\cos ecx=\frac{1}{\sin x}$

$\tan x=\frac{\sin x}{\cos x}$

$\Rightarrow \cot x=\frac{\cos x}{\sin x}$

$\Rightarrow \frac{{\sin }^3\theta }{{\cos }^3\theta \times \left({\sec }^2\theta \right)}+\frac{{\cos }^3\theta }{{\sin }^3\theta \times {\cos ec}^2\theta }-\frac{1}{\cos \theta \sin \theta }+2\sin \theta \cos \theta$

$\Rightarrow \frac{{\sin }^3\theta }{\cos \theta }+\frac{{\cos }^3\theta }{\sin \theta }-\frac{1}{\cos \theta \sin \theta }+2\sin \theta \cos \theta$

$\Rightarrow \frac{{\sin }^4\theta +{\cos }^4\theta -1+2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }$

$\Rightarrow \frac{{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2+2{\sin }^2\theta {\cos }^2\theta -1}{\sin \theta \cos \theta }$

$\because a^2+b^2+2ab={(a+b)}^2$

$\Rightarrow \frac{{({\sin }^2\theta +{\cos }^2\theta )}^2-1}{\sin \theta \cos \theta }$

$\because {\sin }^{2}x+{\cos }^{2}x=1$

$\Rightarrow \frac{1^2-1}{\sin \theta \cos \theta }=0$