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