Question

1/(cosec x + Cot x) - 1/sinx=1/sin x = 1/sin x-1/(cosec x-cot x)

Answer 1

I hope that you can understand all steps....

Answer 2

Answer and Explanation:

To show : $\frac{1}{\csc x+\cot x}-\frac{1}{\sin x} =\frac{1}{\sin x}-\frac{1}{\csc x-\cot x}$

Solution :

Taking LHS,

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

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

$=\frac{(csc x-cot x)(csc x+cot x)}{\csc x+\cot x}-\frac{1}{\sin x}$

$=csc x-cot x-\csc x$

$=-cot x$

Taking RHS,

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

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

$=\frac{1}{\sin x}-\frac{(csc x-cot x)(csc x+cot x)}{\csc x-\cot x}$

$=\csc x-(\csc x+\cot x)$

$=-\cot x$

LHS=RHS

Hence proved.