Question

prove that under root cosec A- 1/ cosecA + 1 + under root cosec A + 1/cosec A - 1= 2 sec A.

Answer 1

We have to prove that,

$\\ \implies \boxed{\sf \sqrt{ \dfrac{ cosec \: A - 1}{cosec \: A + 1} } + \sqrt{ \dfrac{ cosec \: A + 1}{cosec \: A - 1} } = 2 \sec \ A}$

Solving LHS,

$\\ \implies \sf \sqrt{ \dfrac{ cosec \: A - 1}{cosec \: A + 1} \times\dfrac{ cosec \: A - 1}{cosec \: A - 1} } \: + \: \sqrt{ \dfrac{ cosec \: A + 1}{cosec \: A - 1} \times\dfrac{ cosec \: A + 1}{cosec \: A + 1} }$

$\\ \implies \sf \sqrt{ \dfrac{ (cosec \: A - 1) ^{2} }{cosec^{2} \: A - 1 } } \: + \: \sqrt{ \dfrac{ (cosec \: A + 1) ^{2} }{cosec^{2} \: A - 1} }$

$\\ \implies \sf \sqrt{ \dfrac{ (cosec \: A - 1) ^{2} }{ \cot ^{2} A } } \: + \: \sqrt{ \dfrac{ (cosec \: A + 1) ^{2} }{\cot ^{2}A} }$

$\\ \implies \sf { \dfrac{ (cosec \: A - 1) }{ \cot A } } \: + \: { \dfrac{ (cosec \: A + 1) }{\cot A} }$

$\\ \implies \sf { \dfrac{ cosec \: A - 1 + cosec \: A + 1 }{ \cot A } }$

$\\ \implies \sf { \dfrac{ 2 \: cosec \: A }{ \cot A } }$

$\\ \implies \sf { \dfrac{ 2 \: \times \frac{1}{ \sin A} }{ \frac{ \cos A}{ \sin A} } }$

$\\ \implies \sf \dfrac{2}{ \cancel{ \sin A }} \times \dfrac{ \cancel{ \sin A }}{ \cos A }$

$\\ \implies \sf \dfrac{2}{\cos A}$

$\\ \implies \boxed{\sf 2 \sec A}$

• LHS = RHS

HENCE PROVED