Question

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{In}∆ABC: \\ A+B+C=\mathrm{\pi } \\ \therefore A+B=\mathrm{\pi }-C \\ \mathrm{Now},\mathrm{LHS}=\cos \left(A+B\right)+\cos C \\ =\cos \left(\mathrm{\pi }-C\right)+\cos C \\ =-\cos \left(C\right)+\cos C\left[\because \cos \left(\mathrm{\pi }-C\right)=-\cos \left(C\right)\right] \\ =0 \\ =\mathrm{RHS} \\ \mathrm{H}\mathrm{ence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{In}∆ABC: \\ A+B+C=\mathrm{\pi } \\ \Rightarrow A+B=\mathrm{\pi }-C \\ \Rightarrow \frac{A+B}{2}=\frac{\mathrm{\pi }-C}{2} \\ \Rightarrow \frac{A+B}{2}=\frac{\mathrm{\pi }}{2}-\frac{C}{2} \\ \mathrm{Now},\mathrm{LHS}=\cos \left(\frac{A+B}{2}\right) \\ =\cos \left(\frac{\mathrm{\pi }}{2}-\frac{C}{2}\right) \\ =\sin \left(\frac{C}{2}\right)\left[\because \cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)=\sin \mathrm{\theta }\right] \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{In}∆ABC: \\ A+B+C=\mathrm{\pi } \\ \Rightarrow A+B=\mathrm{\pi }-C \\ \Rightarrow \frac{A+B}{2}=\frac{\mathrm{\pi }-C}{2} \\ \Rightarrow \frac{A+B}{2}=\frac{\mathrm{\pi }}{2}-\frac{C}{2} \\ \mathrm{Now},\mathrm{LHS}=\tan \left(\frac{A+B}{2}\right) \\ =\tan \left(\frac{\mathrm{\pi }}{2}-\frac{C}{2}\right) \\ =\cot \left(\frac{C}{2}\right)\left[\because \tan \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\cot \mathrm{\theta }\right] \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$