A) \[\sin A\]
B) \[\cos A\]
C) \[\tan A\]
D) \[\sec A\]
Correct Answer: B
Solution :
[b] \[(1+\sin A)(\sec A-\tan A)=(1+\sin A)\,\,\left( \frac{1}{\cos A}-\frac{\sin A}{\cos A} \right)\] |
\[=(1+\sin A)\frac{(1-\sin A)}{\cos A}=\frac{1-{{\sin }^{2}}A}{\cos A}=\frac{{{\cos }^{2}}A}{\cos A}=\cos A\] |
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