If frac{sin A - sin C}{cos C - cos A} = cot B | vedclass

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

Question

(A) Given: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$Using the sum-to-product formulas:$\sin A - \sin C = 2 \cos \frac{A + C}{2} \sin \frac{A

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(A) Given: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$Using the sum-to-product formulas:$\sin A - \sin C = 2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}$$\cos C - \cos A = 2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}$Substituting these into the equation:$\frac{2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}}{2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}} = \cot B$$\cot \frac{A + C}{2} = \cot B$$\frac{A + C}{2} = B$$A + C = 2B$This condition implies that $A, B, C$ are in $A.P.$

If $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$,then $A, B, C$ are in

Similar Questions

If the angles of a triangle are in the ratio $1: 1: 4$,then the ratio of the perimeter of the triangle to its largest side is

In triangle $ABC$,if $a=4, b=3, c=2$,then $2(a-b \cos C)(a-c \sec B) = $

In a triangle $ABC$,if $a=5, b=3, c=7$,then $\sqrt{\frac{\sin(A-B)}{\sin(A+B)}}=$

In a triangle $ABC$,if $\sin A \sin B = \frac{ab}{c^2}$,then the triangle is

In $\triangle ABC$,$\angle B = \frac{\pi}{4}$ and $\angle C = \frac{\pi}{3}$. If the area of the triangle is $54 + 18\sqrt{3}$ sq. units,then $a =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module