Question

√1+sinA/1-sinA+√1-sinA/1+sinA =2(sinA*tanA+cosA​)

Answer 1

Answer:

If is a common question in Trigonometry , we can found this in many books and exams.

Here is the solution -

=(sinA + 1 - cosA) / (cosA - 1 + sinA)

Divide each term by cosA

=(tanA + secA - 1) / (1 - secA + tanA)

=[tanA + secA - ( sec2A−tan2A )] /

( 1 - secA + tanA)

= (tanA+secA)( 1 - secA + tanA )/(1 - secA + tanA)

= tanA + secA

= (sinA / cosA )+ 1 / cosA

= (sinA + 1 )/ cosA

Hence Proved

Hope it will help you