Question

tanA+tanB/cotA+cotB=tanAtanB prove

Answer 1

(tan A+tan B) / (cot A+cot B)
⇒(tan A+tan B) / (1/tan A+1/tan B)
⇒(tan A +tan B) / (tan A+tan B / tan A·tan B)
⇒(tan A+tan B) / (tan A·tan B / tan A+tan B)
⇒tan A ·tan B
hence proved

Answer 2

To prove:

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

Let’s consider the LHS,

LHS $=\frac{\tan A+\tan B}{\cot A+\cot B}$

$\Rightarrow \frac{\tan A+\tan B}{\frac{1}{\tan A}+\frac{1}{\tan B}}$               $\left[{since}, \cot \theta=\frac{1}{\tan \theta}\right]$

$\Rightarrow \frac{\tan A+\tan B}{\frac{\tan A+\tan B}{\tan A \tan B}}$

$\Rightarrow \tan A \tan B$

= RHS

Hence proved.

To prove LHS = RHS, we used some trigonometric equations. Since cot is the inverse of tan. We substituted it in the equations and easily proves that left hand side is equal to right hand side.