Question

If $\frac{2 sin\theta}{(1 + cos\theta + sin\theta)}$ = x, then find the value of $\frac{(1 - cos\theta + sin\theta)}{(1 + sin\theta)}$

Answer 1

Given, $\frac{2 sin\theta}{(1 + cos\theta + sin\theta)}$ = x

or, $\frac{2sin\theta}{1+cos\theta+sin\theta}\times\frac{1-cos\theta+sin\theta}{1-cos\theta+sin\theta}=x$

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

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

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

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

hence, $\frac{(1 - cos\theta + sin\theta)}{(1 + sin\theta)}$ = x

Answer 2

HELLO DEAR,

Given,
$\bold{\frac{2 sin\Theta}{(1 + cos\Theta + sin\Theta)}}$ = x

=> $\bold{\frac{2sin\Theta}{1+cos\Theta + sin\Theta} * \frac{1-cos\Theta + sin\Theta}{1-cos\Theta + sin\Theta}}$ = x

=> $\bold{\frac{2sin\Theta(1-cos\Theta + sin\Theta )}{(1+sin\Theta)^2 -cos^2\theta}}$ = x

=> $\bold{\frac{2sin\Theta(1-cos\Theta+sin\Theta)}{1+sin^2\Theta+2sin\Theta-cos^2\Theta}}$ = x

=> $\bold{\frac{2sin\Theta(1-cos\Theta+sin\Theta)}{2sin^2\Theta+2sin\Theta)}}$ = x

=> $\bold{\frac{1-cos\Theta +sin\Theta}{1+sin\Theta}}$ = x

hence, $\bold{\frac{(1 - cos\Theta + sin\Theta)}{(1 + sin\Theta)}}$ = x

I HOPE IT'S HELP YOU DEAR,
THANKS