Prove the following trigonometric identities: √1+sinA/1-sinA = secA + tanA

Question

Prove the following trigonometric identities:

\(\sqrt{\frac{1+sinA}{1-sinA}}\) = secA + tanA

Answer 1

\(\sqrt{\frac{1+sinA}{1-sinA}}\) = \(\sqrt{\frac{1+sinA}{1-sinA}}\times \sqrt{\frac{1+sinA}{1+sinA}}\) 

  =   \(\sqrt{\frac{(1+sinA)^2}{1-sin^2A}}\)  

= \(\sqrt{\frac{(1+sinA)^2}{cos^2A}}\)  

=  \(\frac{1+sinA}{cosA}\)  

=  \(\frac{1}{cosA}+\frac{sinA}{cosA}\)

= secA + tanA

Hence Proved.