Question

If $A,B,C$ be the angles of a triangle, then $\sum \frac{\cot A+\cot B}{\tan A+\tan B}$

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step 1: Find the value of $\sum \frac{\cot A+\cot B}{\tan A+\tan B}$

Given that, $A,B,C$ are the angles of a triangle.

$⟹A+B+C=180°$

$⟹\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\begin{array}{rl}\frac{\cot A+\cot B}{\tan A+\tan B}& =\frac{\frac{1}{\tan A}+\frac{1}{\tan B}}{\tan A+\tan B}\\ & =\frac{\frac{(\tan B+\tan A)}{\tan A\tan B}}{\tan A+\tan B}\\ & =\frac{1}{\tan A\tan B}\end{array}$

Step 2: Find the value of $\sum \frac{\cot A+\cot B}{\tan A+\tan B}$

$\begin{array}{cc}\sum \frac{\cot A+\cot B}{\tan A+\tan B}& =\frac{1}{\tan A\tan B}+\frac{1}{\tan B\tan C}+\frac{1}{\tan C\tan A}\\ & =\frac{\tan C+\tan A+\tan B}{\tan A\tan B\tan C}\\ & =\frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}\\ & =\frac{\tan A\tan B\tan C}{\tan A\tan B\tan C}\\ & =1\end{array}$

Hence the value of $\sum \frac{\cot A+\cot B}{\tan A+\tan B}$ is $1$.

Hence option $\left(A\right)$ is the correct option.