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

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

Question

(B) Given that $A+B+C=180^{\circ}$.We can write $A+B = 180^{\circ}-C$.Dividing both sides by $2$,we get $\frac{A+B}{2} = \frac{180^{\circ}-C}{2} = 90^

Vedclass · Accepted Answer

$\cot \frac{C}{2}$

Vedclass · Answer

(B) Given that $A+B+C=180^{\circ}$.We can write $A+B = 180^{\circ}-C$.Dividing both sides by $2$,we get $\frac{A+B}{2} = \frac{180^{\circ}-C}{2} = 90^{\circ}-\frac{C}{2}$.Now,taking the tangent on both sides: $	an \left(\frac{A+B}{2}ight) = 	an \left(90^{\circ}-\frac{C}{2}ight)$.Using the trigonometric identity $	an(90^{\circ}-	heta) = \cot 	heta$,we get $	an \left(\frac{A+B}{2}ight) = \cot \frac{C}{2}$.

If $A+B+C=180^{\circ},$ then $\tan \left(\frac{A+B}{2}\right)=$ ...........

Similar Questions

$\cos 35^{\circ} = \ldots \ldots \ldots$

If $\sin \theta - \cos \theta = 0$,then the value of $(\sin^4 \theta + \cos^4 \theta)$ is

If $\cot \theta = \frac{a}{b}$,then $\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = \ldots$

If $\tan \theta = \frac{5}{12}$,then $\cos \theta = \ldots$

$\sin^{2} 15^{\circ} + \sin^{2} 75^{\circ} = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module