In any triangle ABC,(a + b)^2 sin^2 frac{C}{2 | vedclass

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

Question

(A) Expanding the expression: $(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2}$ $= (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2} + (a^2 + b^2 - 2ab

Vedclass · Accepted Answer

$c^2$

Vedclass · Answer

(A) Expanding the expression: $(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2}$ $= (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2} + (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2}$ $= a^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + b^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$ $= a^2 + b^2 - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ $= a^2 + b^2 - 2ab \cos C$ Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$. Therefore,the expression equals $c^2$.

In any triangle $ABC$,$(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2} =$

Similar Questions

If $a, b$ and $c$ are the sides of a triangle such that $a^4 + b^4 + c^4 = 2c^2(a^2 + b^2)$,then the angle opposite to the side $c$ is:

In $\triangle ABC$,if $r_1=8, r_2=12$ and $r_3=24$,then the ordered triple $(a, b, c) =$

With the usual notations,in a $\triangle ABC$,if $a=2, b=\sqrt{6}$ and $c=\sqrt{3}+1$,then $\sin^2 C - \sin^2 A =$

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

If the sides of a triangle are in the ratio $\sqrt{3} : \sqrt{5} : \sqrt{8+\sqrt{15}}$,then the largest angle in that triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module