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

(C) Given $A + B + C = 180^o$,we have $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.Taking $\cot$ on both sides: $\cot(\frac{A}{2} + \frac{B}{2}) =

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $A + B + C = 180^o$,we have $\frac{A}{2} + \frac{B}{2} = 90^o - \frac{C}{2}$.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 $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = 	an \frac{C}{2}$.Since $	an \frac{C}{2} = \frac{1}{\cot \frac{C}{2}}$,we have $\frac{\cot \frac{A}{2} \cot \frac{B}{2} - 1}{\cot \frac{A}{2} + \cot \frac{B}{2}} = \frac{1}{\cot \frac{C}{2}}$.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,we get $\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} - \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2}$.Rearranging the terms,we find $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$.

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 maximum value of $(\cos \alpha_1) \cdot (\cos \alpha_2) \ldots (\cos \alpha_n)$ under the constraints $0 \leq \alpha_1, \alpha_2, \ldots, \alpha_n \leq \frac{\pi}{2}$ and $(\cot \alpha_1) \cdot (\cot \alpha_2) \ldots (\cot \alpha_n) = 1$ is

The maximum value of $(7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$ for every $\theta \in R$.

If $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $,then the difference between the maximum and minimum values of ${u^2}$ is given by

Prove that $\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$

If $A, B, C$ are the angles of a triangle,then $\sin 2A - \sin 2B + \sin 2C =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module