A) \[\sin 3A\]
B) \[\cos 3A\]
C) \[\sin A+\cos A\]
D) 3
Correct Answer: D
Solution :
\[\frac{{{\sin }^{3}}A+\sin 3A}{\sin A}+\frac{{{\cos }^{3}}-\cos 3A}{\cos A}\] \[\Rightarrow \frac{{{\sin }^{3}}A+\sin A-4{{\sin }^{3}}A}{\sin A}+\frac{{{\cos }^{3}}A-\left[ 4{{\cos }^{3}}A-3\cos A \right]}{\cos A}\]\[\Rightarrow \frac{3\sin A-3{{\sin }^{3}}A}{\sin A}+\frac{\left( -3{{\cos }^{3}}A+3\cos A \right)}{\cos A}\] \[=3-3{{\sin }^{2}}A-3{{\cos }^{2}}A+3\] \[=6-3({{\cos }^{2}}A+{{\sin }^{2}}A)\] \[=6-3(1)\] \[=3\]You need to login to perform this action.
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