In a ∆ABC, prove that:
(i) cos (A + B) + cos C = 0
(ii) $\cos \left(\frac{A+B}{2}\right)=\sin \frac{C}{2}$
(iii) $\tan \frac{A+B}{2}=\cot \frac{C}{2}$
(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.
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