\[\frac{\tan A+\tan B}{\cot A+\cot B}\]is equal to [SSC (CGL) 2014] |
A) \[\cot A\cot B\]
B) \[\sec A\,\,\text{cosec B}\]
C) \[\tan A\tan B\]
D) None of these
Correct Answer: C
Solution :
\[\frac{\tan A+\tan B}{\cot A+\cot B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}\] |
\[=\frac{\frac{\sin A\cdot \cos B+\cos A\cdot \sin B}{\cos A\cdot \cos B}}{\frac{\cos A\cdot \sin B+\sin A\cdot \cos B}{\sin A\cdot \sin B}}\] |
\[=\frac{\sin A\cdot \sin B}{\cos A\cdot \cos B}=\tan A\cdot \tan B\] |
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