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If $\large\frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}$$=x$ then $\large\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}$=?

$(a)\;\large\frac{1}{x}$$\qquad(b)\;x\qquad(c)\;1-x\qquad(d)\;1+x$

1 Answer

Given :$x=\large\frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}$
$\Rightarrow x=\large\frac{2\sin\alpha}{(1+\sin \alpha)+(\cos\alpha)}\frac{1+\sin\alpha-\cos\alpha}{(1+\sin\alpha)-\cos\alpha}$
$\Rightarrow \large\frac{2\sin\alpha(1-\cos\alpha+\sin\alpha)}{(1+\sin\alpha)^2-\cos^2\alpha}$
$\Rightarrow \large\frac{2\sin\alpha(1-\cos\alpha+\sin\alpha)}{1+\sin^2\alpha+2\sin\alpha-\cos^2\alpha}$
$\Rightarrow \large\frac{2\sin\alpha(1-\cos\alpha+\sin\alpha)}{2\sin\alpha(1+\sin\alpha)}$
$\Rightarrow \large\frac{1-\cos \alpha+\sin\alpha}{1+\sin\alpha}$
$\Rightarrow \large\frac{1-\cos \alpha+\sin\alpha}{1+\sin\alpha}$$=x$
Hence (b) is the correct option.
answered Oct 10, 2013 by sreemathi.v
 

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