In △ABC, prove that-

**snigdha** · February 2nd 2010, 08:00 AM

1. tan (B+C)=-tan A

2. cos (B+C)/2 = sin A/2

**Grandad** · February 2nd 2010, 10:45 AM

Hello snigdha

Originally Posted by snigdha

1. tan (B+C)=-tan A

2. cos (B+C)/2 = sin A/2

First, note that:

$\tan(180^o-\theta) = -\tan\theta$

and

$\cos(90^o-\theta) = \sin\theta$

1.

$A+B+C = 180^o$

$B+C = 180^o-A$

$\Rightarrow \tan(B+C) = \tan(180^o-A)$

$=-\tan A$

2.

$B+C = 180^o-A$

$\Rightarrow \tfrac12(B+C) = 90^o-\tfrac12A$

$\Rightarrow \cos\tfrac12(B+C) = \cos(90^o-\tfrac12A)$

$=\sin\tfrac12A$

Grandad

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