A) \[{{\cos }^{2}}A\]
B) \[{{\cos }^{2}}B\]
C) \[{{\sin }^{2}}A\]
D) \[{{\sin }^{2}}B\]
Correct Answer: C
Solution :
We know that, \[2s=a+b+c\] \[\therefore \frac{\left( a+b+c \right)\left( b+c-a \right)\left( c+a-b \right)\left( a+b-c \right)}{4{{b}^{2}}{{c}^{2}}}\] \[=\frac{2s\left( 2s-2a \right)\left( 2s-2b \right)\left( 25-2c \right)}{4{{b}^{2}}{{c}^{2}}}\] \[=4\frac{s(s-a)}{bc}\times \frac{(s-b)(s-c)}{bc}\] \[=4{{\cos }^{2}}\frac{A}{2}\times {{\sin }^{2}}\frac{A}{2}\] \[={{\sin }^{2}}A\]
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