If A + B + C = π,

Question:

If $A+B+C=\pi$, then $\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$ is equal to

(a) tan A tan B tan C

(b) 0

(c) 1

(d) None of these

Solution:

(c) 1

π = 180°

Using tan(180 – A) = -tan A, we get:

$C=\pi-(A+B)$

Now,

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

$=\frac{\tan A+\tan B+\tan [\pi-(A+B)]}{\tan A \tan B \tan [\pi-(A+B)]}$

 

$=\frac{\tan A+\tan B-\tan (A+B)}{-\tan A \tan B t \operatorname{an}(A+B)}$

$=\frac{\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A \tan B}}{-\tan A \tan B \times \frac{\tan A+\tan B}{1-\tan A \tan B}}$

$=\frac{\tan A+\tan B-\tan ^{2} A \tan B-\tan A \tan ^{2} B-\tan A-\tan B}{-\tan ^{2} A \tan B-\tan A \tan ^{2} B}$

$=\frac{-\tan ^{2} A \tan B-\tan A \tan ^{2} B}{-\tan ^{2} A \tan B-\tan A \tan ^{2} B}$

 

$=1$

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