In a ΔABC a+b+cb+c−ac+a−ba+b−c4b2c2=___

a+b+cb+c−ac+a−ba+b−c4b2c2=2s2s−a−a2s−b−b2a−c−c4b2c2=16ss−as−bs−c4b2c2=4Δ2b2c2=sin2A∵Δ=12bc sinA

In a ΔABC a+b+cb+c−ac+a−ba+b−c4b2c2=___

a+b+cb+c−ac+a−ba+b−c4b2c2=2s2s−a−a2s−b−b2a−c−c4b2c2=16ss−as−bs−c4b2c2=4Δ2b2c2=sin2A∵Δ=12bc sinA

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In a $Δ A B C$ $\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} =___$

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$\sin^{2} A$

$\cos^{2} B$

$\cos^{2} A$

$\sin^{2} B$

answer is C.

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Detailed Solution

$\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}}$

$= \frac{2 s (2 s - a - a) (2 s - b - b) (2 a - c - c)}{4 b^{2} c^{2}}$

$= \frac{16 s (s - a) (s - b) (s - c)}{4 b^{2} c^{2}} = \frac{4 Δ^{2}}{b^{2} c^{2}} = \sin^{2} A$

$[∵ Δ = \frac{1}{2} b c \sin A]$

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In a ΔABC a+b+cb+c−ac+a−ba+b−c4b2c2=___

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In a $Δ A B C$ $\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} =___$