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

The value of $\cot(7.5^{\circ}) + \tan(67.5^{\circ}) - \cot(67.5^{\circ}) - \tan(7.5^{\circ})$ is:

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

The value of $\cos ^{2} \theta+\sec ^{2} \theta$ is always

If $y = (1 + \tan A)(1 - \tan B)$ where $A - B = \frac{\pi}{4}$,then $(y + 1)^{y + 1}$ is equal to

If $\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y,$ then $\frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module