If cos⁡(θ−α)=a and cos⁡(θ−β)=b then sin2⁡(α−β)+2abcos⁡(α−β) is equal to

A
$a^{2} + b^{2}$
B
$a^{2} - b^{2}$
C
$b^{2} - a^{2}$
D
$- a^{2} - b^{2}$

Solution:

$\begin{array}{l} ∵ α - β = (θ - β) - (θ - α) \\ \begin{array}{ll} ∴ & \cos (α - β) = \cos ((θ - β) - (θ - α)) = \cos (θ - β) \cos (θ - α) + \sin (θ - β) \sin (θ - α) \end{array} \\ and \sin (α - β) = \sin (θ - β) \cos (θ - α) + \cos (θ - β) \sin (θ - α) \\ \Rightarrow \cos (α - β) = b \cdot a + \sqrt{1 - a^{2}} \sqrt{1 - b^{2}} . . . . . . . . . . . . . . . . . (1) \\ and \sin (α - β) = (a \sqrt{1 - b^{2}} - b \sqrt{1 - a^{2}}) \\ Now, \sin^{2} (α - β) = {((a \sqrt{1 - b^{2}}) + (b \sqrt{1 - a^{2}}))}^{2} \\ \Rightarrow \sin^{2} (α - β) = a^{2} (1 - b^{2}) + b^{2} (1 - a^{2}) - 2 ab {\cos (α - β) - ab} f r o m e q (1) \\ \Rightarrow \sin^{2} (α - β) + 2 abcos (α - β) = a^{2} - a^{2} b^{2} + b^{2} - b^{2} a^{2} + 2 a^{2} b^{2} \\ \Rightarrow \sin^{2} (α - β) + 2 abcos (α - β) = a^{2} + b^{2} \end{array}$

If $cos⁡(θ−α)=a$ and $\cos (θ - β) = b$ then $\sin^{2} (α - β) + 2 abcos (α - β)$ is equal to

Solution:

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