Question

Prove that: $\frac{1}{\csc \theta -\cot \theta }-\frac{1}{\sin \theta }=\frac{1}{\sin \theta }-\frac{1}{\csc \theta +\cot \theta }$

Answer 1

$L.H.S$

$\frac{1}{\csc \theta +\cot \theta }-\frac{1}{\sin \theta }$

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

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

$\csc \theta -\cot \theta -\csc \theta$

$-\cot \theta$

$R.H.S$

$\frac{1}{\sin \theta }-\frac{1}{\csc \theta -\cot \theta }$

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

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

$\csc \theta -\csc \theta -\cot \theta$

$-\cot \theta$

$\therefore LHS=RHS$

Hence, proved.