In a triangle ABC,frac{a-b}{a+b} = | vedclass

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

Question

(C) Using the sine rule,we have $a = k \sin A$ and $b = k \sin B$. Substituting these into the expression: $\frac{a-b}{a+b} = \frac{k \sin A - k \sin

Vedclass · Accepted Answer

$	an \left(\frac{A-B}{2}ight) 	an \frac{C}{2}$

Vedclass · Answer

(C) Using the sine rule,we have $a = k \sin A$ and $b = k \sin B$. Substituting these into the expression: $\frac{a-b}{a+b} = \frac{k \sin A - k \sin B}{k \sin A + k \sin B} = \frac{\sin A - \sin B}{\sin A + \sin B}$ Using the sum-to-product formulas: $= \frac{2 \cos \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)}{2 \sin \left(\frac{A+B}{2}ight) \cos \left(\frac{A-B}{2}ight)}$ $= \cot \left(\frac{A+B}{2}ight) 	an \left(\frac{A-B}{2}ight)$ Since $A+B+C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$. Therefore,$\cot \left(\frac{A+B}{2}ight) = \cot \left(90^{\circ} - \frac{C}{2}ight) = 	an \frac{C}{2}$. Thus,$\frac{a-b}{a+b} = 	an \frac{C}{2} 	an \left(\frac{A-B}{2}ight)$.

In a $\triangle ABC$,$\frac{a-b}{a+b} = $

Similar Questions

In a triangle $ABC$,with usual notations,if $b=\sqrt{3}, c=1, \angle A=30^{\circ}$,then angle $B$ is (in $^{\circ}$)

In a triangle $ABC$,with usual notations,if $3b = a + c$,then $\cot \frac{A}{2} \cdot \cot \frac{C}{2} = $

In $\Delta ABC$,if $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$,then the triangle is:

In a $\Delta ABC,$ if $b = 20, c = 21$ and $\sin A = 3/5$,then $a = $

In a $\triangle ABC$,the expression $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module