Question

Prove that $\frac{\sin \theta }{\cot \theta +\csc \theta }=2+\frac{\sin \theta }{\cot d\theta -\csc \theta }$

Answer 1

We have,

$\frac{\sin \theta }{\cot \theta +\csc \theta }=2+\frac{\sin \theta }{\cot \theta -\csc \theta }$

$\frac{\sin \theta }{\cot \theta +\csc \theta }-\frac{\sin \theta }{\cot \theta -\csc \theta }=2$ …….. (1)

Consider the L.H.S

$=\frac{\sin \theta }{\cot \theta +\csc \theta }-\frac{\sin \theta }{\cot \theta -\csc \theta }$

$=\frac{\sin \theta (\cot \theta -\csc \theta )-\sin \theta (\cot \theta +\csc \theta )}{(\cot \theta +\csc \theta )(\cot \theta -\csc \theta )}$

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

$=\frac{\cos \theta -1-\cos \theta -1}{({\cot }^2\theta -{\csc }^2\theta )}$

$=\frac{-2}{-1}[\because {\cot }^{2}x-{\csc }^{2}x=-1]$

$=2$

R.H.S

Hence, proved.