In triangle ABC, frac{1 + cos(A - B)cos C}{1 | vedclass

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

Question

(D) We know that in a triangle $ABC,$ $C = 180^\circ - (A + B),$ so $\cos C = -\cos(A + B).$Substituting this into the numerator: $1 + \cos(A - B)\cos

Vedclass · Accepted Answer

$\frac{a^2 + b^2}{a^2 + c^2}$

Vedclass · Answer

(D) We know that in a triangle $ABC,$ $C = 180^\circ - (A + B),$ so $\cos C = -\cos(A + B).$Substituting this into the numerator: $1 + \cos(A - B)\cos C = 1 - \cos(A - B)\cos(A + B).$Using the identity $\cos(A - B)\cos(A + B) = \cos^2 A - \sin^2 B,$ we get $1 - (\cos^2 A - \sin^2 B) = 1 - \cos^2 A + \sin^2 B = \sin^2 A + \sin^2 B.$Similarly,for the denominator: $1 + \cos(A - C)\cos B = 1 - \cos(A - C)\cos(A + C) = 1 - (\cos^2 A - \sin^2 C) = \sin^2 A + \sin^2 C.$Thus,the expression becomes $\frac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C}.$Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k,$ we have $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}.$Substituting these values,we get $\frac{(a/k)^2 + (b/k)^2}{(a/k)^2 + (c/k)^2} = \frac{a^2 + b^2}{a^2 + c^2}.$

In triangle $ABC,$ $\frac{1 + \cos(A - B)\cos C}{1 + \cos(A - C)\cos B} = $

Similar Questions

In $\triangle ABC$,if the line joining the circumcentre $(O)$ and the incentre $(I)$ is parallel to $BC$,then $\cos B + \cos C = $

In a $\triangle ABC$,the expression $\frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b}$ is equal to

In a $\Delta ABC,$ if $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)},$ then $a^2, b^2, c^2$ are in

In a $\triangle ABC$,$\operatorname{cosec} A(\sin B \cos C + \cos B \sin C)$ is equal to

In a triangle $ABC$,with usual notations,$2 ac \sin \left(\frac{A-B+C}{2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module