In a triangle ABC,tan frac{A}{2} tan frac{B}{ | vedclass

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

Question

(B) In a triangle $ABC$,$A+B+C = \pi$. Dividing by $2$,we get $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A}{2} + \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) In a triangle $ABC$,$A+B+C = \pi$. Dividing by $2$,we get $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}$. Taking tangent on both sides: $	an(\frac{A}{2} + \frac{B}{2}) = 	an(\frac{\pi}{2} - \frac{C}{2}) = \cot \frac{C}{2}$. Using the identity $	an(x+y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we have $\frac{	an \frac{A}{2} + 	an \frac{B}{2}}{1 - 	an \frac{A}{2} 	an \frac{B}{2}} = \frac{1}{	an \frac{C}{2}}$. Cross-multiplying gives: $	an \frac{C}{2}(	an \frac{A}{2} + 	an \frac{B}{2}) = 1 - 	an \frac{A}{2} 	an \frac{B}{2}$. Rearranging the terms: $	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} = 1$.

In a triangle $ABC$,$\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} =$

Similar Questions

In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)

If in a triangle $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$,then the triangle is

If $A = 30^\circ, c = 7\sqrt{3}$ and $C = 90^\circ$ in $\Delta ABC$,then $a = $

In a triangle $ABC$,with usual notations,if $a=5, b=4$,and $\cos(A-B)=\frac{31}{32}$,then $c=$

In $\triangle ABC$,if $a^2 \sin^2 \frac{C}{2} + c^2 \sin^2 \frac{A}{2} = \frac{b^2}{2}$,then $a+c : b =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module