1. Tardigrade
  2. Question
  3. Mathematics
  4. In Δ ABC (a-b)2 cos2 (C/2) + (a+b)2 sin2 (C/2) =

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. In $\Delta ABC \left(a-b\right)^{2} cos^{2} \frac{C}{2} + \left(a+b\right)^{2} sin^{2} \frac{C}{2} = $

MHT CETMHT CET 2016Trigonometric Functions

Solution:

$(a-b)^{2} cos ^{2} \frac{C}{2}+(a+b)^{2} sin ^{2} \frac{C}{2}$
$=\left(a^{2}+b^{2}-2 ab\right) cos ^{2} \frac{C}{2}+a^{2}+b^{2}+2ab sin ^{2} \frac{C}{2}$
$=a^{2}\left(cos ^{2} \frac{C}{2}+sin ^{2} \frac{C}{2}\right)+b^{2}\left(cos ^{2} \frac{C}{2}+\sin ^{2} \frac{C}{2}\right)$
$-2 ab\left(cos ^{2} \frac{C}{2}-sin ^{2} \frac{C}{2}\right)$
$=a^{2}+b^{2}-2 a b\,cos\, C$
$\left[\because cos ^{2} \frac{C}{2}-\sin ^{2} \frac{C}{2}=\cos C\right]$
$=a^{2}+b^{2}-2 a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)=c^{2}$
$\left[\because \cos C=\frac{a^{2}+b^{2}}{2ab}\right]$