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

$\cos 2\theta + 2\cos \theta$ is always

For all values of $\theta$,the values of $3-\cos \theta+\cos \left(\theta+\frac{\pi}{3}\right)$ lie in the interval :

For $\theta > \frac{\pi}{3}$,the value of $f(\theta) = \sec^2 \theta + \cos^2 \theta$ always lies in the interval

Let $y=4 \sin^2 \theta - \cos 2 \theta$. If $l$ and $m$ are the minimum and maximum values of $y$ respectively,then

Find the maximum and minimum values of the function given by $h(x) = \sin(2x) + 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module