Question

If $x=\frac{2sin\theta }{1+cos\theta +sin\theta }$, then prove that $\frac{1-cos\theta +sin\theta }{1+sin\theta }$ is also equal to x.

Answer 1

$\frac{2sin\theta }{1+cos\theta +sin\theta }=x$

$\Rightarrow \frac{2sin\theta (1-cos\theta +sin\theta )}{(1+cos\theta +sin\theta )(1-cos\theta +sin\theta )}=x\left[Rationalizingthedenominator\right]$

$\Rightarrow \frac{2sin\theta (1-cos\theta +sin\theta )}{(1+sin\theta {)}^2-co{s}^2\theta }=x$

$\Rightarrow \frac{2sin\theta -2sin\theta cos\theta +2si{n}^2\theta }{1+si{n}^2\theta +2sin\theta -co{s}^2\theta }=x$

$\Rightarrow \frac{2sin\theta (1+cos\theta -sin\theta )}{2si{n}^2\theta +2sin\theta }=x$

$\Rightarrow \frac{2sin\theta (1+cos\theta -sin\theta )}{2sin\theta (1+sin\theta )}=x$

$\Rightarrow \frac{1+cos\theta -sin\theta }{1+sin\theta }=x$

[Taking $2sin\theta$ common form numerator and denominator]Hence Proved.