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  4. If y = (2 sin α/1 + cos α + sin α) then ( 1 - cos α + sin α /1 + sin α) is equal to

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Q. If $ y = \frac{2 \sin \, \alpha}{1 + \cos \, \alpha + \sin \, \alpha}$ then $\frac{ 1 - \cos \alpha + \sin \, \alpha }{1 + \sin \alpha}$ is equal to

Trigonometric Functions

Solution:

Now, $\frac{1-\cos \alpha + \sin \alpha}{1 +\sin \alpha}$
$ = \frac{\left(1- \cos\alpha + \sin\alpha\right) \left(1+\cos\alpha + \sin\alpha\right)}{\left(1+\sin\alpha\right)\left(1+\cos \alpha + \sin\alpha\right)} $
$= \frac{\left(1+\sin \alpha\right)^{2} -\cos^{2}\alpha}{\left(1+\sin \alpha\right)\left(1+\cos\alpha + \sin\alpha\right) } $
$= \frac{1+\sin^{2} \alpha+ 2\sin \alpha-\left(1 - \sin^{2} \alpha\right)}{\left(1+\sin\alpha\right)\left(1+\cos\alpha + \sin\alpha\right)} $
$=\frac{1+\sin ^{2} \alpha+2 \sin \alpha-1+\sin ^{2} \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{2 \sin \alpha+2 \sin ^{2} \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}$
$=y$