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# The value of $\sin(n+1)A\sin(n+2)A+\cos(n+1)A\cos(n+2)A$ is equal to

$(a)\;\sin A\qquad(b)\;\cos A\qquad(c)\;-\cos A\qquad(d)\;\sin 2A$

Can you answer this question?

We have
$\sin(n+1)A\sin(n+2)A+\cos(n+1)A\cos(n+2)A$
$\Rightarrow \cos(n+2)A\cos(n+1)A+\sin(n+2)A\sin(n+1)A$
$\Rightarrow \cos\{(n+2)A-(n+1)A\}$
$\Rightarrow \cos A$
Hence (b) is the correct answer.
answered Oct 4, 2013