Question

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

Answer 1

L.H.S$=\frac{1}{\csc A-\cot A}-\frac{1}{\sin A}$

$=\frac{1}{\csc A-\cot A}\times \frac{\csc A+\cot A}{\csc A+\cot A}-\frac{1}{\sin A}$

$=\frac{\csc A+\cot A}{{\csc }^{2}A-{\cot }^{2}A}-\frac{1}{\sin A}$

$=\frac{\csc A+\cot A}{1}-\csc A$

$=\csc A+\cot A-\csc A$

$=\cot A$ .......$\left(1\right)$

R.H.S$=\frac{1}{\sin A}-\frac{1}{\csc A+\cot A}$

$=\frac{1}{\sin A}-\frac{1}{\csc A+\cot A}\times \frac{\csc A-\cot A}{\csc A-\cot A}$

$=\csc A-\frac{\csc A-\cot A}{{\csc }^{2}A-{\cot }^{2}A}$

$=\csc A-\frac{\csc A-\cot A}{1}$

$=\csc A-\csc A+\cot A$

$=\cot A$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have L.H.S$=$R.H.S

Hence proved.