Question

In √1+sin A /1-sin A = sec A + tan A, Prove RHS=LHS.​

Answer 1

Answer:

${ \huge{ \underline { \pmb{ \sf{Required \: Solution:}}}}}$

$: { \implies{ \sf{ \sqrt{ \frac{1 + sinA}{1 - sinA} } }}} \\ \\ { \rm{Rationalizing \: with \: Denominator}} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{1 + sinA}{1 - sinA} \times \frac{1 + sinA}{1 + sinA} } }}} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{ {(1)}^{2} - {sin}^{2}A } } }}} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{1 - {sin}^{2} A} } }}} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{ {cos}^{2} A} } }}} \\ \\ Square \: and \: root \: got \:cancelled \\ \\: { \implies{ \sf{ \frac{1 + sinA}{cosA} }}} \\ \\ : { \implies{ \sf{ \frac{1}{cosA} + \frac{sinA}{cosA} }}} \\ \\ : { \implies{ \sf{secA + tanA}}} \\ \\ hence \: proved$

Used identity:

• Sin²A + Cos²A = 1
• Cos²A = 1 - Sin²A