If A + B + C = 180^{circ},then tan A + tan B | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $A + B + C = 180^{\circ}$,we have $A + B = 180^{\circ} - C$. Taking tangent on both sides: $	an(A + B) = 	an(180^{\circ} - C)$. Using the

Vedclass · Accepted Answer

$	an A 	an B 	an C$

Vedclass · Answer

(C) Given $A + B + C = 180^{\circ}$,we have $A + B = 180^{\circ} - C$. Taking tangent on both sides: $	an(A + B) = 	an(180^{\circ} - C)$. Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(180^{\circ} - C) = -	an C$,we get: $\frac{	an A + 	an B}{1 - 	an A 	an B} = -	an C$. Multiplying both sides by $(1 - 	an A 	an B)$,we get: $	an A + 	an B = -	an C(1 - 	an A 	an B)$. $	an A + 	an B = -	an C + 	an A 	an B 	an C$. Rearranging the terms,we get: $	an A + 	an B + 	an C = 	an A 	an B 	an C$.

If $A + B + C = 180^{\circ}$,then $\tan A + \tan B + \tan C = $

Similar Questions

The minimum value of $27^{\cos 2x} 81^{\sin 2x}$ is.......

If $A+B+C=\pi$ and $\cos B=\cos A \cos C$,then $\tan A \tan C=$

If $A, B, C$ are acute positive angles such that $A + B + C = \pi$ and $\cot A \cot B \cot C = K$,then:

For $a \in R - \{0\}$,if $a \cos x + a \sin x + a = 2K + 1$ has a solution,then $K$ lies in the interval

$cosec^2 \theta = \frac{4xy}{(x + y)^2}$ is true if and only if

Vedclass Test Series

Exam Paper Generator

Online Exam Module