If $x=\frac{2 sin θ}{(1+cos θ+sin θ)}$, then the value of $\frac{1−cos θ+sin θ}{(1+sin θ)}$ is:
Answer & explanation
Correct answer: option 2
We are given that :-
x = \(\frac{ 2 sinθ}{ 1 + cosθ + sinθ }\)
Now multiply and divide RHS by ( 1 - cosθ + sinθ )
= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1 + cosθ + sinθ) × ( 1 - cosθ + sinθ )}\)
= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)² - cos²θ}\)
{ we know , sin²θ + cos²θ = 1 }
= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)² - ( 1 - sin²θ) }\)
= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)² - ( 1 - sinθ).( 1+ sinθ) }\)
= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)(1 + sinθ - 1+ sinθ) }\)
= \(\frac{ ( 2 sinθ ) × ( 1 - cosθ + sinθ ) }{( 1+ sinθ)(2 sinθ) }\)
= \(\frac{ ( 1 - cosθ + sinθ ) }{( 1+ sinθ) }\)
So,
\(\frac{ ( 1 - cosθ + sinθ ) }{( 1+ sinθ) }\) = x