South Asian Economic Union
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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.
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