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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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If $\large\frac{2\sin \alpha}{1+\cos \alpha+\sin \alpha}$$=y$,then prove that $\large\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$ is equal to $y$

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$\large\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$
$\large\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}\frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}$
$\large\frac{1-(\cos \alpha-\sin \alpha)}{1+\sin \alpha}\frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}$
$\large\frac{1+\cos \alpha+\sin \alpha-\cos \alpha+\sin \alpha-\big[(\cos \alpha-\sin \alpha)(\cos \alpha+\sin \alpha)\big]}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$\large\frac{1+\cos \alpha+\sin \alpha-\cos \alpha+\sin \alpha-\big[(\cos^2 \alpha-\sin^2 \alpha)\big]}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$\large\frac{1+2\sin \alpha-\cos^2\alpha+\sin^2\alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$\large\frac{2\sin \alpha+2\sin^2\alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$\large\frac{2\sin \alpha (1+\sin \alpha)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}$
$\large\frac{2\sin \alpha}{1+\cos \alpha+\sin \alpha}$$=y$
Hence proved.
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