Question

Prove the following identities (1-17)

$1-\frac{{\sin }^2\mathrm{\theta }}{1+\cot \mathrm{\theta }}-\frac{{\cos }^2\mathrm{\theta }}{1+\tan \mathrm{\theta }}=\sin \mathrm{\theta }\cos \mathrm{\theta }$

Answer 1

$1-\frac{{\sin }^2\mathrm{\theta }}{1+\cot \mathrm{\theta }}-\frac{{\cos }^2\mathrm{\theta }}{1+\tan \mathrm{\theta }}=\sin \mathrm{\theta }\cos \mathrm{\theta }$


$$\begin{gathered} \mathrm{LHS}=1-\frac{{\sin }^3\theta }{\sin \theta +\cos \theta }-\frac{{\cos }^3\theta }{\sin \theta +\cos \theta } \\ =\frac{\sin \theta +\cos \theta -\left({\sin }^3\theta +{\cos }^3\theta \right)}{\sin \theta +\cos \theta } \\ =\frac{\left(\sin \theta +\cos \theta \right)\left(1-{\sin }^2\theta -{\cos }^2\theta +\sin \theta \cos \theta \right)}{\sin \theta +\cos \theta } \\ =\left(1-{\sin }^2\theta -{\cos }^2\theta +\sin \theta \cos \theta \right) \\ =\left(1-1+\sin \theta \cos \theta \right) \\ =\sin \theta \cos \theta \end{gathered}$$

= RHS
Hence proved.