Question

please answer this two questions I request you​

Answer 1

1. We have LHS

$\large\implies \bf \: \dfrac{sec \:A - tan \:A + 1 }{sec \: A - tan \: A - 1}$

We know that :-

➠sec²A-tan²A+1

$\large\implies \bf \: \dfrac{sec \:A - tan \:A + ( {sec}^{2} A - {tan}^{2} A) }{sec \: A - tan \: A - ( {sec}^{2} A - {tan}^{2} A)}$

$\large\implies \bf \: \dfrac{ \cancel{sec \:A - tan \:A} (1+ {sec}A - {tan}A) }{ \cancel{sec \: A - tan \: A }(1- {sec}A - {tan}A)}$

$\large\implies \bf \: \dfrac{ (1+ {sec}A - {tan}A) }{(1- {sec}A - {tan}A)} = RHS$

2. In second question also we will do same process as follows :-

We have LHS

$\large\implies \bf \: \dfrac{1 - cosec \:A + cot\:A }{1 + cosec \: A - tan \: A }$

We know that

➠cosec²A-tan²A= 1 , so by putting the value of 1 in above we will get the result

$\large\implies \bf \: \dfrac{cosec^{2}A - {tan}^{2} A- cosec \:A + cot\:A }{cosec^{2} A - {tan}^{2} A+ cosec \: A - tan \: A }$

$\large\implies \bf \: \dfrac{ \cancel{(cosec \: A - cot \: A})cosec \:A+cot\:A - 1}{ \cancel{(cosec \: A - cot \:A) }cosec \: A + tan \: A + 1}$

$\large\implies \bf \: \dfrac{ cosec \:A+cot\:A - 1}{ cosec \: A + tan \: A + 1} = RHS$

Hence, proved