Discussion Forum

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

Answer:

Explanation:

$(a-b)^{2}\cos^{2}\frac{C}{2}+(a+b)^{2}\sin^{2}\frac{C}{2}$

=  $(a^{2}+b^{2}-2ab)\cos^{2}\frac{C}{2}+(a^{2}+b^{2}+2ab)\sin^{2}\frac{C}{2}$

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

 = $a^{2}+b^{2}-2ab \cos C$

 $\left[\because\cos^{2}\frac{C}{2}-\sin^{2}\frac{C}{2}=\cos C\right]$

     =   $a^{2}+b^{2}-2ab \left(\frac{a^{2}+b^{2}-c^{2}}{2ab}\right)=c^{2}$

                                                                             $\left[\because \cos C =\frac{a^{2}+b^{2}}{2ab}\right]$