Question

(Tan beta - tan alpha) / tan alpha tan beta

Answer 1

To prove that: \dfrac{\tan \alpha+\tan \beta}{\cot \alpha +\cot \beta} =\tan \alpha\tan \beta.

Solution:

L.H.S = \dfrac{\tan \alpha+\tan \beta}{\cot \alpha +\cot \beta}

= \dfrac{\tan \alpha+\tan \beta}{\dfrac{1}{\tan \alpha} +\dfrac{1}{\tan \beta} }

Using the trigonometric identity:

\cot A = \dfrac{1}{\tan A}

Taking LCM of denominator part, we get

= \dfrac{\tan \alpha+\tan \beta}{\dfrac{\tan \alpha+\tan \beta}{\tan \alpha\tan \beta} }

= \tan \alpha\tan \beta

= R.H.S., proved.

Thus, \dfrac{\tan \alpha+\tan \beta}{\cot \alpha +\cot \beta} =\tan \alpha\tan \beta, proved.