If A + B + C = 180^o,then the value of cot fr | vedclass

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

Question

(A) Given $A + B + C = 180^o$,so $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^o$.This implies $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.Taking

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $A + B + C = 180^o$,so $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^o$.This implies $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.Taking $\cot$ on both sides: $\cot(\frac{A}{2} + \frac{B}{2}) = \cot(90^o - \frac{C}{2})$.Using the formula $\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$,we get $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = 	an \frac{C}{2} = \frac{1}{\cot \frac{C}{2}}$.Cross-multiplying gives: $(\cot \frac{A}{2} \cot \frac{B}{2} - 1) \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.Expanding this: $\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} - \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.Rearranging the terms: $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$.

If $A + B + C = 180^o$,then the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$ is equal to

Similar Questions

If $\alpha$ is a root of $25\cos^2\theta + 5\cos\theta - 12 = 0$ and $\pi/2 < \alpha < \pi$,then $\sin 2\alpha$ is equal to

The value of $\cos 24^{\circ} + \cos 55^{\circ} + \cos 125^{\circ} + \cos 204^{\circ} + \cos 300^{\circ}$ is

Evaluate: $\tan A + \cot (180^\circ + A) + \cot (90^\circ + A) + \cot (360^\circ - A)$

$\sin 36^\circ \sin 72^\circ \sin 108^\circ \sin 144^\circ = $

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module