A) \[{{\cos }^{2}}A\]
B) \[si{{n}^{2}}A\]
C) \[cos\,\,A\,\,\cos \,\,A\,\,\cos \,\,C\]
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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