In ABC, prove that ( b - c )^2 ^2 ( A 2 )+( b + c )^2 ^2 ( A 2 )= a ^2

L.H.S. =(b-c)^2 ^2 A 2 +(b+c)^2 ^2 A 2 = (b^2+c^2-2 b c ) ^2 a 2 + (b^2+c^2+2 b c ) ^2 A 2 = (b^2+c^2 ) ^2 A 2 -2 b c ^2 A 2 + (b^2+c^2 ) ^2 A 2 +2 b c ^2 A 2 = (b^2+c^2 ) ( ^2 A 2 + ^2 A 2 )-2 b c ( ^2 A 2 - ^2 A 2 ) = (b^2+c^2 )(1)-2 b c A [ ^2 - ^2 = 2 ] =b^2+c^2-2 b c A =a^2 . .[ By cosine rule ] = R.H.S.

In ABC, prove that ( b - c )^2 ^2 ( A 2 )+( b + c )^2 ^2 ( A 2 )=…

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Question

In $\triangle ABC$, prove that $( b - c )^2 \cos ^2\left(\frac{ A }{2}\right)+( b + c )^2 \sin ^2\left(\frac{ A }{2}\right)= a ^2$

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Answer

$\text { L.H.S. }=(b-c)^2 \cos ^2 \frac{A}{2}+(b+c)^2 \sin ^2 \frac{A}{2}$
$=\left(b^2+c^2-2 b c\right) \cos ^2 \frac{a}{2}+\left(b^2+c^2+2 b c\right) \sin ^2 \frac{A}{2}$
$=\left(b^2+c^2\right) \cos ^2 \frac{A}{2}-2 b c \cos ^2 \frac{A}{2}+\left(b^2+c^2\right) \sin ^2 \frac{A}{2}+2 b c \sin ^2 \frac{A}{2}$
$=\left(b^2+c^2\right)\left(\cos ^2 \frac{A}{2}+\sin ^2 \frac{A}{2}\right)-2 b c\left(\cos ^2 \frac{A}{2}-\sin ^2 \frac{A}{2}\right)$
$=\left(b^2+c^2\right)(1)-2 b c \cos A \ldots \ldots\left[\because \cos ^2 \theta-\sin ^2 \theta=\cos 2 \theta\right]$
$=b^2+c^2-2 b c \cos A$
$=a^2 \quad \ldots \ldots . .[\text { By cosine rule }]$
$=\text { R.H.S. }$

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