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 a triangle $ABC$ the sides $AB$ and $AC$ are perpendicular,then the true equation is

In a $\triangle ABC$,if $a=6$,$b=5$,and $c=4$,then find the value of $\cos 2A$.

In a $\triangle ABC$,the expression $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to:

The smallest angle of the triangle whose sides are $6 + \sqrt{12}$,$\sqrt{48}$,and $\sqrt{24}$ is:

If $a^2, b^2, c^2$ are in $A.P.$,then which of the following are also in $A.P.$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module