sinθ/1+cosθ is equal to

Question

\(\frac{sinθ}{1+cosθ}\)is equal to

A. \(\frac{1+cosθ}{sinθ}\)

B. \(\frac{1- cosθ}{cosθ}\)

C. \(\frac{1- cosθ}{sinθ}\)

D. \(\frac{1- sinθ}{cosθ}\)

Answer 1

To find: \(\frac{sinθ}{1+cosθ}\)

Consider \(\frac{sinθ}{1+cosθ}\)

Rationalizing the above fraction by (1 – cos θ),

\(\frac{sinθ}{1+cosθ}\) = \(\frac{sinθ}{1+cosθ}\times \frac{1-cosθ}{1-cosθ}\) 

= \(\frac{sinθ(1-cosθ)}{(1+cosθ)(1-cosθ)}\)

= \(\frac{sinθ(1-cosθ)}{(1-cos^2θ)} \) [∵ (a – b) (a + b) = a² – b²]

∵ sin²θ + cos²θ = 1 

⇒ sin²θ = 1 – cos²θ