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If $\sin\alpha+\sin\beta=\alpha$ and $\cos\alpha+\cos\beta=b$ then $\sin(\alpha+\beta)$ is equal to

$(a)\;\large\frac{2ab}{a^2-b^2}\qquad$$(b)\;\large\frac{ab}{a^2-b^2}\qquad$$(c)\;\large\frac{ab}{a^2+b^2}$$\qquad(d)\;\large\frac{2ab}{a^2+b^2}$

1 Answer

Now $\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
We have $\cos(\alpha+\beta)=\large\frac{b^2-a^2}{b^2+a^2}$
$\Rightarrow \sqrt{1-\bigg(\large\frac{b^2-a^2}{b^2+a^2}\bigg)^2}$
$\Rightarrow \sqrt{\large\frac{4a^2b^2}{(a^2+b^2)^2}}$
$\Rightarrow \large\frac{2ab}{b^2+a^2}$
Hence (d) is the correct answer.
answered Oct 14, 2013 by sreemathi.v
 

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