Question

In a \Delta ABC, prove that:

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

$\left(ii\right)cos\left(\frac{A+B}{2}\right)=sin\frac{C}{2}$

$\left(iii\right)tan\frac{A+B}{2}=cot\frac{C}{2}$

Answer 1

(i)We have A+B+C $=\pi [\because$ sum of 3 angles of a triangle is $\pi ={180}^{\circ }]$

$\Rightarrow A+B=\pi -C$

$\Rightarrow cos(A+B)=cos(\pi -C)\Rightarrow =-cosC[\because (\pi -\theta )=-cos\theta ]$

$\Rightarrow cos(A+B)+cosC=0$ Hence proved.

(ii)We have A+B+C $=\pi [\because$ sum of 3 angles of a triangle is $\pi ={180}^{\circ }]$

$\Rightarrow A+B=\pi -C$

$\Rightarrow \frac{A+B}{2}=\frac{\pi -C}{2}$

$\Rightarrow \frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$

$\Rightarrow cos=\left(\frac{A+B}{2}\right)=cos(\frac{\pi }{2}-\frac{C}{2})$

$\Rightarrow sin\frac{C}{2}$

$[\because cos(\frac{\pi }{2}-\theta )=sin\theta ]$

Hence $cos\left(\frac{A+B}{2}\right)=sin\frac{C}{2}$ Proved.

(iii)We have A+B+C$=\pi [\because$ sum of 3 angles of a triangle is $\pi ={180}^{\circ }]$

$\Rightarrow A+B=\pi -C$

$\Rightarrow \frac{A+B}{2}=\frac{\pi -C}{2}$

$\Rightarrow \frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$

$\Rightarrow tan=\left(\frac{A+B}{2}\right)=tan(\frac{\pi }{2}-\frac{C}{2})$

$\Rightarrow cot\frac{C}{2}$

$[\because tan(\frac{\pi }{2}-\theta )=cot\theta ]$

Hence $tan\left(\frac{A+B}{2}\right)=cot\frac{C}{2}$ Hence proved.