In a $△ ABC, \frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}}$ equals

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$\cos^{2} A$

$\sin^{2} A$

$\sin^{2} B$

$\cos^{2} B$

answer is C.

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Detailed Solution

$\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} = \frac{2 s 2 (s - a) 2 (s - b) 2 (s - c)}{4 b^{2} c^{2}} = \frac{4 Δ^{2}}{b^{2} c^{2}} = \frac{4 {(\frac{1}{2} bcsinA)}^{2}}{b^{2} c^{2}} = \sin^{2} A$

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