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

Let $\Delta$ denote the area of a $\triangle ABC$. If $\alpha, \beta, \gamma$ are the lengths of the altitudes of the $\triangle ABC$,then $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$

In a $\triangle ABC$,the altitudes from $B$ and $C$ to the opposite sides are not shorter than their respective opposite sides. Then,one of the angles of $\triangle ABC$ is $........^{\circ}$

In any $\triangle ABC$,the expression $\frac{1+\cos(A-B) \cdot \cos C}{1+\cos(A-C) \cdot \cos B}$ is equal to:

If in a triangle $ABC$, $\sin A, \sin B, \sin C$ are in $A.P.$, then

In a triangle $ABC$,if $(b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right) = K(1 - \cos 2A)$,then $K =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module