A) 1
B) 2
C) 4
D) None of these
Correct Answer: A
Solution :
We have \[\sin A,\,\cos A\] and \[\tan A\] are in G.P. \[{{\cos }^{2}}A=\sin A\,\tan A=\frac{{{\sin }^{2}}A}{\cos A}\,\Rightarrow \,{{\cos }^{3}}A-{{\sin }^{2}}A=0\] Hence \[{{\cos }^{3}}A+{{\cos }^{2}}A={{\sin }^{2}}A+{{\cos }^{2}}A=1\]You need to login to perform this action.
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