In a triangle ABC,if a neq b,then the value o | vedclass

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

Question

(A) Given,in triangle $ABC$,we need to evaluate $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$. Using the sine rule,$a = 2R \sin A$ and $b

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given,in triangle $ABC$,we need to evaluate $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$. Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$. Substituting these values: $= \frac{2R \sin A \cos A - 2R \sin B \cos B}{2R \sin A \cos B - 2R \sin B \cos A} + \cos C$ $= \frac{\sin 2A - \sin 2B}{2(\sin A \cos B - \sin B \cos A)} + \cos C$ Using the formula $\sin 2A - \sin 2B = 2 \sin(A - B) \cos(A + B)$ and $\sin(A - B) = \sin A \cos B - \sin B \cos A$: $= \frac{2 \sin(A - B) \cos(A + B)}{2 \sin(A - B)} + \cos C$ $= \cos(A + B) + \cos C$ Since $A + B = \pi - C$,we have $\cos(A + B) = \cos(\pi - C) = -\cos C$. $= -\cos C + \cos C = 0$.

In a triangle $ABC$,if $a \neq b$,then the value of $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$ is:

Similar Questions

In $\triangle ABC$,if $a, b, c$ are in arithmetic progression and $C = 2A$,then $a: c =$

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides of the triangle are in

If $\Delta = a^2 - (b - c)^2$ is the area of the $\triangle ABC$,then $\tan A$ is equal to

In a $\triangle ABC$,$\frac{2(r_1+r_3)}{ac(1+\cos B)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module