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If $\cos (θ - α) = a, \cos (θ - β) = b$ , then $\sin^{2} (α - β) + 2 abcos (α - β) =$

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a² + b²

a² - b²

b²-a²

-a²- b²

answer is A.

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Detailed Solution

$\begin{array}{l} \cos (α - β) = \cos [(θ - β) - (θ - α)] \\ = ba + \sqrt{1 - a^{2}} \sqrt{1 - b^{2}} \\ [\cos (α - β) - ab]^{2} = (1 - a^{2}) (1 - b^{2}) \\ \Rightarrow a^{2} + b^{2} = \sin^{2} (α - β) + 2 abcos (α - β) \end{array}$

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