If $A + B = 60 °$ then $4 (\sin^{2} A + \sin^{2} B + \sin A \sin B) =$

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answer is 3.

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Detailed Solution

A + B = 60 °

$\Rightarrow \cos [A + B] = \cos 60 °$

$\cos A . \cos B - \sin A . \sin B = \frac{1}{2}$

$2 \cos A . \cos B = 1 + 2 \sin A . \sin B$

Squares of both sides

$\Rightarrow 4 \cos^{2} A . \cos^{2} B = 1 + 4 \sin^{2} A . \sin^{2} B + 4 \sin A . \sin B$

$\Rightarrow 4 [1 - \sin^{2} A] [1 - \sin^{2} B] = 1 + 4 \sin^{2} A . \sin^{2} B + 4 \sin A . \sin B$

$\Rightarrow 4 [1 - \sin^{2} B - \sin^{2} A + \sin^{2} A . \sin^{2} B] = 1 + 4 \sin^{2} A . \sin^{2} B + 4 \sin A . \sin B$

$\Rightarrow 4 - 4 \sin^{2} B - 4 \sin^{2} B + 4 \sin^{2} A . \sin^{2} B = 1 + 4 \sin^{2} A . \sin^{2} B + 4 \sin A . \sin B$

$\Rightarrow 4 [\sin^{2} B + \sin^{2} A + \sin A . \sin B] = 3$

$\Rightarrow \sin^{2} A + \sin^{2} B + \sin A . \sin B = \frac{3}{4}$

$∴ 4 (\sin^{2} A + \sin^{2} B + \sin A \sin B) = 4 \times \frac{3}{4} = 3$

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