In a Delta ABC,the expression frac{cos C + co | vedclass

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

Question

(B) Given expression: $\frac{\cos C + \cos A}{c + a} + \frac{\cos B}{b}$ Taking the common denominator $b(c + a)$,we get: $= \frac{b(\cos C + \cos A)

Vedclass · Accepted Answer

$\frac{1}{b}$

Vedclass · Answer

(B) Given expression: $\frac{\cos C + \cos A}{c + a} + \frac{\cos B}{b}$ Taking the common denominator $b(c + a)$,we get: $= \frac{b(\cos C + \cos A) + \cos B(c + a)}{b(c + a)}$ $= \frac{b \cos C + b \cos A + c \cos B + a \cos B}{b(c + a)}$ Using the projection formula,we know that $b \cos C + c \cos B = a$ and $b \cos A + a \cos B = c$. Substituting these values: $= \frac{(b \cos C + c \cos B) + (b \cos A + a \cos B)}{b(c + a)}$ $= \frac{a + c}{b(c + a)}$ $= \frac{1}{b}$.

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

Similar Questions

If in the $\Delta ABC$,$AB = 2BC$,then find the ratio $\tan \frac{B}{2} : \cot \left( \frac{C - A}{2} \right)$.

In triangle $ABC$,if $a=7, b=8, \cos C=\frac{2}{7}$ and $C$ is an acute angle,then $c=$

With usual notations, in $\triangle ABC$, $(b - c)^2 \cos^2 \frac{A}{2} + (b + c)^2 \sin^2 \frac{A}{2} = $

In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)

If the angles of a $\triangle ABC$ are in $AP$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module