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If $\cos(\theta-\alpha)=a$ and $\sin(\theta-\beta)=b$ then $\cos^2(\alpha-\beta)+2ab\sin(\alpha-\beta)$ is equal to

$(a)\;4a^2b^2\qquad(b)\;a^2-b^2\qquad(c)\;a^2+b^2\qquad(d)\;-a^2b^2$

1 Answer

$\cos^2(\alpha-\beta)+2ab\sin(\alpha-\beta)$
$\Rightarrow \cos^2(\alpha-\beta)+2\cos(\theta-\alpha)\sin(\theta-\beta)\sin(\alpha-\beta)$
$\Rightarrow \cos^2(\alpha-\beta)+[\sin(2\theta-(\alpha+\beta))+\sin(\alpha-\beta)]\sin(\alpha-\beta)$
$\Rightarrow 1+\sin(2\theta-(\alpha+\beta))\sin(\alpha-\beta)$
$\Rightarrow 1+\sin\{(\theta-\alpha)+(\theta-\beta)\}\sin\{(\theta-\beta)-(\theta-\alpha)\}$
$\Rightarrow 1+\sin^2(\theta-\beta)-\sin^2(\theta-\alpha)$
$\Rightarrow \cos^2(\theta-\alpha)+\sin^2(\theta-\beta)$
$\Rightarrow a^2+b^2$
Hence (c) is the correct answer.
answered Oct 15, 2013 by sreemathi.v
 

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