Question

prove that under root cosec A+1÷Cosec A-1 - under root cosecA -1÷Cosec A+1 =2 cot A
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Answer 1

Correct question:

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

Solving LHS,

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

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

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

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

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

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

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

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

• LHS = RHS

HENCE PROVED