First slide

Trigonometric transformations

Question

If $cosec A + \sec A = cosec B + \sec B$ then $\tan A \tan B$ is equal to

Moderate

A

$\tan (\frac{A + B}{2})$
B

$\cot (\frac{A + B}{2})$
C

$\cot (\frac{A - B}{2})$
D

$\tan (\frac{A - B}{2})$

Solution

$\begin{array}{l} \sec A - \sec B = cosec B - cosec A \\ \Rightarrow \frac{\cos B - \cos A}{\cos A \cos B} = \frac{\sin A - \sin B}{\sin A \sin B} \\ \Rightarrow \frac{\sin A \sin B}{\cos A \cos B} = \frac{\sin A - \sin B}{\cos B - \cos A} \\ \Rightarrow \tan A \tan B = \frac{2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})}{2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2})} \\ = \cot (\frac{A + B}{2}) \end{array}$

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