A) \[{{\cos }^{2}}A\]
B) \[{{\sin }^{2}}A\]
C) \[\cos A\cos B\cos C\]
D) none of these
Correct Answer: B
Solution :
| \[\frac{2s\,(2s-2a)\,\,(2s-2b)\,\,(2s-2c)}{4{{b}^{2}}{{c}^{2}}}\] |
| \[=\frac{16\,(s)(s-a)\,\,(s-b)\,\,(s-c)}{4{{b}^{2}}{{c}^{2}}}=\frac{4{{\Delta }^{2}}}{{{b}^{2}}{{c}^{2}}}\] |
| \[=\frac{4\times {{\left[ \frac{1}{2}bc.\sin A \right]}^{2}}}{{{b}^{2}}{{c}^{2}}}={{\sin }^{2}}A\] |
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