In a ∆ABC, prove that:

Solution:

(i) In $\Delta A B C$ :

$A+B+C=\pi$

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

Now, LHS $=\cos (A+B)+\cos C$

$=\cos (\pi-C)+\cos C$

$=-\cos (C)+\cos C \quad[\because \cos (\pi-C)=-\cos (C)]$

$=0$

= RHS

Hence proved.

(ii) In $\Delta A B C$ :

$A+B+C=\pi$

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

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

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

Now, $\mathrm{LHS}=\cos \left(\frac{A+B}{2}\right)$

$=\cos \left(\frac{\pi}{2}-\frac{C}{2}\right)$

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

= RHS

Hence proved.

(iii) In $\Delta A B C$ :

$A+B+C=\pi$

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

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

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

Now, LHS $=\tan \left(\frac{A+B}{2}\right)$

$=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right)$

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

= RHS

Hence proved.