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Q)

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

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

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A)
$\cos^2(\alpha-\beta)+2ab\sin(\alpha-\beta)=\cos^2(\alpha-\beta)+2\cos(\theta-\alpha)\sin(\theta-\beta)\sin(\alpha-\beta)$
$\qquad\qquad\qquad\qquad\qquad\quad=\cos^2(\alpha-\beta)+[\sin(2\theta-(\alpha+\beta)^2+\sin(\alpha-\beta)]\sin(\alpha-\beta)$
$\qquad\qquad\qquad\qquad\qquad\quad=1+\sin\big(2\theta-(\alpha+\beta)\big)\sin(\alpha-\beta)$
$\qquad\qquad\qquad\qquad\qquad\quad=1+\sin\{(\theta-\alpha)+(\theta-\beta)\}\sin\{(\theta-\beta)-(\theta-\alpha)\}$
$\qquad\qquad\qquad\qquad\qquad\quad=1+\sin^2(\theta-\beta)-\sin^2(\theta-\alpha)$
$\qquad\qquad\qquad\qquad\qquad\quad=\cos^2(\theta-\alpha)+\sin^2(\theta-\beta)$
$\qquad\qquad\qquad\qquad\qquad\quad=a^2+b^2$
Hence (c) is the correct answer.
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