A) \[{{a}^{2}}\]
B) \[{{b}^{2}}\]
C) \[{{c}^{2}}\]
D) None of these
Correct Answer: C
Solution :
\[({{a}^{2}}+{{b}^{2}}-2ab){{\cos }^{2}}\frac{C}{2}+({{a}^{2}}+{{b}^{2}}+2ab){{\sin }^{2}}\frac{C}{2}\] \[={{a}^{2}}+{{b}^{2}}+2ab\left( {{\sin }^{2}}\frac{C}{2}-{{\cos }^{2}}\frac{C}{2} \right)\] \[={{a}^{2}}+{{b}^{2}}-2ab\cos C={{a}^{2}}+{{b}^{2}}-({{a}^{2}}+{{b}^{2}}-{{c}^{2}})={{c}^{2}}\].
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