Question

√1+sinA÷1-sinA = secA+tanA​

Answer 1

Answer:

correct question

I think my answer will help you

Answer 2

Answer:

Correct Question:

• ${ \sf{ \sqrt{ \frac{1 + sinA}{1 - sinA} } }}$

Solution:

$: { \implies{ \sf{ \sqrt{ \frac{1 + sinA}{1 - sinA} } }}} \\ \\ { \rm{Rationalizing \: With \: Denominator}} \\ \\ : { \implies{ \sf{{ \sqrt{ \frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}}}}}} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{ {(1)}^{2} - {sin}^{2} A} } }}} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{1 - {sin}^{2} A} } }}} \\ \\ { \rm{ {sinA}^{2} + {cos}^{2}A = 1 }} \\ \therefore{ \rm{ {cos}^{2} A = 1 - {sin}^{2}A }} \\ \\ : { \implies{ \sf{ \sqrt{ \frac{ {(1 + sinA)}^{2} }{ {cos}^{2}A } } }}} \\ \\ { \rm{square \: and \: root \: got \: cancelled}} \\ \\ : { \implies{ \sf{ \frac{1 + sinA}{cosA} }}} \\ \\ : { \implies{ \sf{ \frac{1 }{cosA} + \frac{sin A}{cosA} }}} \\ \\ : { \implies{ \sf{secA + tanA}}} \\ \\ \therefore{ \rm{Hence \: Proved}}$