Hello snigdha Originally Posted by
snigdha 1. tan (B+C)=-tan A
2. cos (B+C)/2 = sin A/2
First, note that:$\displaystyle \tan(180^o-\theta) = -\tan\theta$
and$\displaystyle \cos(90^o-\theta) = \sin\theta$
1.$\displaystyle A+B+C = 180^o$
$\displaystyle B+C = 180^o-A$
$\displaystyle \Rightarrow \tan(B+C) = \tan(180^o-A)$ $\displaystyle =-\tan A$
2.$\displaystyle B+C = 180^o-A$
$\displaystyle \Rightarrow \tfrac12(B+C) = 90^o-\tfrac12A$
$\displaystyle \Rightarrow \cos\tfrac12(B+C) = \cos(90^o-\tfrac12A)$ $\displaystyle =\sin\tfrac12A$
Grandad