Question

tan alpha+tan beta/cot alpha +cot beta = tan alpha tan beta

Answer 1

Here is the proof to your question...

Answer 2

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.