Question

सिद्ध करें कि $\frac{\tan A+\tan B}{\cot A+\cot B}=\tan A \tan B$.

Answer 1

Solution : बा॰प० $=\frac{\tan A+\tan B}{\cot A+\cot B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}-\frac{\frac{\sin A \cos B+\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \sin B+\sin A \cos B}{\sin A \sin B}}$

$=\frac{\sin A \cos B+\cos A \sin B}{\cos A \cos B} \times \frac{\sin A \sin B}{\cos A \sin B+\sin A \cos B}$

$=\frac{\sin A \sin B}{\cos A \cos B}=\tan A \tan B=$दा॰प०