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

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to

If the angles of a triangle $ABC$ are in the ratio $1: 2: 3$,then the corresponding sides are in the ratio

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\cos A: \cos B: \cos C =$

In $\triangle ABC$,find the value of $\sqrt{\frac{r r_2}{r_3 r_1}}$.

In $\triangle ABC$,if $(a+c)^2 = b^2 + 3ca$,then $\frac{a+c}{2R} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module