Question

$\huge\underline\mathsf\purple{Question}$

Given:-

Prove:-

$\Rightarrow\Huge\boxed {LHS} = {RHS}$​

Answer 1

Answer :

Note: Check this attachment!

Hence:

$\implies$ $\sf{- cot \: x = - cot \: x}$

$\implies$ $\sf{L. H. S. = R. H. S.}$

Answer 2

→ Take L.H.S

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

$=\dfrac{c s c^{2} x-c o t^{2} x}{\csc x+\cot x}-\dfrac{1}{\sin x}$

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

$=\bf{csc x-cot x-csc x}$

$=\bf{-cotx}$

→ Take R.H.S

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

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

$\dfrac{1}{\sin x}-\dfrac{(\csc x-\cot x)(\csc x+\cot x)}{\cos x-\cot x}$

$=\bf{csc x-(csc x+cot x)}$

$=\bf{-cotx}$

L.H.S = R.H.S

Hence Proved

Regards :)