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 $x + y + z = 180^o,$ then $\cos 2x + \cos 2y - \cos 2z$ is equal to

$\frac{\cot \theta - \operatorname{cosec} \theta + 1}{\cot \theta + \operatorname{cosec} \theta - 1}$ is equal to

$\frac{\sin 81^\circ + \cos 81^\circ}{\sin 81^\circ - \cos 81^\circ}$ is equal to

If $A$ lies in the second quadrant and $3\tan A + 4 = 0,$ the value of $2\cot A - 5\cos A + \sin A$ is equal to

If $\frac{x}{\cos \theta} = \frac{y}{\cos \left( \theta - \frac{2\pi}{3} \right)} = \frac{z}{\cos \left( \theta + \frac{2\pi}{3} \right)},$ then $x + y + z = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module