√ 1+sinθ/1-sinθ is equal to

Question

\(\sqrt{\frac{1+sinθ}{1-sinθ}}\) is equal to 

A. sec θ + tan θ 

B. sec θ − tan θ 

C. sec²θ + tan²θ 

D. sec²θ − tan²θ

Answer 1

Note: Since all the options involve the trigonometric ratios sec θ and tan θ, so we divide the whole term (numerator as well as denominator) by cos θ.

To find: \(\sqrt{\frac{1+sinθ}{1-sinθ}}\)

Consider \(\sqrt{\frac{1+sinθ}{1-sinθ}}\)

Dividing numerator and denominator by cos θ, we get

Rationalizing the term by multiplying it by \(\sqrt{sec θ + tan θ}\),

= \(\sqrt{\frac{(secθ+tanθ)^2}{sec^2θ-tan^2θ}}\)  

Now, as 1 + tan²θ = sec²θ 

⇒ sec²θ – tan²θ = 1

⇒ \(\sqrt{\frac{1+sinθ}{1-sinθ}}\) = \(\sqrt{\frac{(secθ+tanθ)^2}{sec^2θ-tan^2θ}}\) 

= \(\sqrt{(sec θ + tan θ)^2}\)

= secθ – tanθ