Question

1/ cosecx +cotx -1/sinx = 1/sinx -1/cosecx - cotx​

Answer 1

$\displaystyle\huge\green{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

${cosec }^{2}A - {cot }^{2} A= 1$

CALCULATION

LHS

$\displaystyle \: \frac{1}{Sin x} - \frac{1}{ (cosec x + cot x)}$

$= \displaystyle \: {cosec x} - \frac{ (cosec x - cot x)}{ (cosec x + cot x)(cosec x-cot x)}$

$= \displaystyle \:{cosec x} - \frac{ ({cosec } x - cot x)}{ ({cosec }^{2}x - {cot }^{2} x)}$

$= \displaystyle \: {cosec x} - \frac{ ({cosec } x - cot x)}{ 1}$

$= \displaystyle \: {cosec x} - { ({cosec } x - cot x)}$

$= {cot x}$

RHS

$\displaystyle \: \frac{1}{ (cosec x-cot x)}- \frac{1}{Sin x}$

$= \displaystyle \: \frac{ (cosec x + cot x)}{ (cosec x + cot x)(cosec x-cot x)}- {cosec x}$

$= \displaystyle \: \frac{ ({cosec } x+ cot x)}{ ({cosec }^{2}x - {cot }^{2} x)}- {cosec x}$

$= \displaystyle \: \frac{ ({cosec } x + cot x)}{ 1}- {cosec x}$

$= \displaystyle \: { ({cosec } x + cot x)}- {cosec x}$

$= cot x$

Hence LHS = RHS

Hence proved