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

$\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ} = \ldots \ldots \ldots \ldots$

Prove that,
$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta$

Write 'True' or 'False' and justify your answer.
$(\tan \theta+2)(2 \tan \theta+1)=5 \tan \theta+\sec ^{2} \theta$

Prove that,
$1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha} = \operatorname{cosec} \alpha$

$(1+\tan ^{2} \theta)(1-\cos ^{2} \theta) = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module