Prove that:$ \frac{\cot A+\tan B}{\cot B+\tan A}=\cot A \tan B $

To do:

We have to prove that \( \frac{\cot A+\tan B}{\cot B+\tan A}=\cot A \tan B \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$\frac{\cot A+\tan B}{\cot B+\tan A}=\frac{\frac{\cos A}{\sin A}+\frac{\sin B}{\cos B}}{\frac{\cos B}{\sin B}+\frac{\sin A}{\cos A}}$

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

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

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

$=\cot A \tan B$

Hence proved.     

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Updated on: 10-Oct-2022

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