11. In a triangle ABC, prove that (a) cos(A+B)+cos C=0 (b) tan (A+B)÷ 2= cot C÷ 2

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Δ A B C

a (b cos C - c cos B) = b^{2} - c^{2}

\tan (\frac{C + A}{2}) = \cot \frac{B}{2}

\sin (\frac{B + C}{2}) = \cos \frac{A}{2}

A B C,

tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}

A B C,

tan \frac{A - B}{2} = \frac{a - b}{b + a} cot \frac{C}{2}

In a \Delta ABC, prove that:

(i)cos (A+B)+cos C =0

$(i i) c o s (\frac{A + B}{2}) = s i n \frac{C}{2}$

$(i i i) t a n \frac{A + B}{2} = c o t \frac{C}{2}$

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