Question

Prove that:

(i) $\frac{\sin \left(A+B\right)+\sin \left(A-B\right)}{\cos \left(A+B\right)+\cos \left(A-B\right)}=\tan A$
(ii) $\frac{\sin \left(A-B\right)}{\cos A\cos B}+\frac{\sin \left(B-C\right)}{\cos B\cos C}+\frac{\sin \left(C-A\right)}{\cos C\cos A}=0$
(iii) $\frac{\sin \left(A-B\right)}{\sin A\sin B}+\frac{\sin \left(B-C\right)}{\sin B\sin C}+\frac{\sin \left(C-A\right)}{\sin C\sin A}=0$
(iv) sin² B = sin² A + sin² (A − B) − 2 sin A cos B sin (A − B)
(v) cos² A + cos² B − 2 cos A cos B cos (A + B) = sin² (A + B)
(vi) $\frac{\tan \left(A+B\right)}{\cot \left(A-B\right)}=\frac{{\tan }^{2}A-{\tan }^{2}B}{1-{\tan }^{2}A{\tan }^{2}B}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=\frac{\sin \left(A+B\right)+\sin \left(A-B\right)}{\cos \left(A+B\right)+\cos \left(A-B\right)} \\ =\frac{\sin A\cos B+\cos A\sin B+\sin A\cos B-\cos A\sin B}{\cos A\cos B-\sin A\sin B+\cos A\cos B+\sin A\sin B} \\ =\frac{2\sin A\cos B}{2\cos A\cos B} \\ =\frac{\sin A}{\cos A} \\ =\tan A \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}=\frac{\sin \left(A-B\right)}{\cos A\cos B}+\frac{\sin \left(B-C\right)}{\cos B\cos C}+\frac{\sin \left(C-A\right)}{\cos C\cos A} \\ =\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}+\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}+\frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A} \\ =\frac{\sin A\cos B}{\cos A\cos B}-\frac{\cos A\sin B}{\cos A\cos B}+\frac{\sin B\cos C}{\cos B\cos C}-\frac{\cos B\sin C}{\cos B\cos C}+\frac{\sin C\cos A}{\cos C\cos A}-\frac{\cos C\sin A}{\cos C\cos A} \\ =\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}+\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}+\frac{\sin C}{\cos C}-\frac{\sin A}{\cos A} \\ =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{LHS}=\frac{\sin \left(A-B\right)}{\sin A\sin B}+\frac{\sin \left(B-C\right)}{\sin B\sin C}+\frac{\sin \left(C-A\right)}{\sin C\sin A} \\ =\frac{\sin A\cos B-\cos A\sin B}{\sin A\sin B}+\frac{\sin B\cos C-\cos B\sin C}{\sin B\sin C}+\frac{\sin C\cos A-\cos C\sin A}{\sin C\sin A} \\ =\frac{\sin A\cos B}{\sin A\sin B}-\frac{\cos A\sin B}{\sin A\sin B}+\frac{\sin B\cos C}{\sin B\sin C}-\frac{\cos B\sin C}{\sin B\sin C}+\frac{\sin C\cos A}{\sin C\sin A}-\frac{\cos C\sin A}{\sin C\sin A} \\ =\frac{\cos B}{\sin B}-\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}-\frac{\cos B}{\sin B}+\frac{\cos A}{\sin A}-\frac{\cos C}{\sin C} \\ =cotB-cotA+cotC-cotB+cotA-cotC \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{RHS}={\sin }^{2}A+{\sin }^2\left(A-B\right)-2\sin A\cos B\sin \left(A-B\right) \\ ={\sin }^{2}A+\sin \left(A-B\right)\left\{\sin \left(A-B\right)-2\sin A\cos B\right\} \\ ={\sin }^{2}A+\sin \left(A-B\right)\left(\sin A\cos B-\cos A\sin B-2\sin A\cos B\right) \\ ={\sin }^{2}A+\sin \left(A-B\right)\left(-\sin A\cos B-\cos A\sin B\right) \\ ={\sin }^{2}A-\sin \left(A-B\right)\left(\sin A\cos B+\cos A\sin B\right) \\ ={\sin }^{2}A-\sin \left(A-B\right)\sin \left(A+B\right) \\ ={\sin }^{2}A-\left({\sin }^{2}A-{\sin }^{2}B\right) \\ ={\sin }^{2}A-{\sin }^{2}A+{\sin }^{2}B \\ ={\sin }^{2}B \\ =\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}={\cos }^{2}A+{\cos }^{2}B-2\cos A\cos B\cos \left(A+B\right) \\ ={\cos }^{2}A+1-{\sin }^{2}B-2\cos A\cos B\cos \left(A+B\right) \\ =1+{\cos }^{2}A-{\sin }^{2}B-2\cos A\cos B\cos \left(A+B\right) \\ =1+{\cos }^{2}A-{\sin }^{2}B-2\cos A\cos B\cos \left(A+B\right) \\ =1+\cos \left(A+B\right)\cos \left(A-B\right)-2\cos A\cos B\cos \left(A+B\right) \\ =1+\cos \left(A+B\right)\left\{\cos \left(A-B\right)-2\cos A\cos B\right\} \\ =1+\cos \left(A+B\right)\left(\cos A\cos B+\sin A\sin B-2\cos A\cos B\right) \\ =1+\cos \left(A+B\right)\left(-\cos A\cos B+\sin A\sin B\right) \\ =1-\cos \left(A+B\right)\left(\cos A\cos B-\sin A\sin B\right) \\ =1-\cos \left(A+B\right)\cos \left(A+B\right) \\ =1-{\cos }^2\left(A+B\right) \\ ={\sin }^2\left(A+B\right) \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{LHS}=\frac{\tan \left(A+B\right)}{cot\left(A-B\right)} \\ =\frac{\tan \left(A+B\right)}{{\displaystyle \frac{1}{\tan \left(A-B\right)}}} \\ =\tan \left(A+B\right)\times \tan \left(A-B\right) \\ =\frac{\tan A+\tan B}{1-\tan A\tan B}\times \frac{\tan A-\tan B}{1+\tan A\tan B} \\ =\frac{\left(\tan A+\tan B\right)\left(\tan A-\tan B\right)}{\left(1-\tan A\tan B\right)\left(1+\tan A\tan B\right)} \\ =\frac{{\left(\tan A\right)}^2-{\left(\tan B\right)}^2}{{\left(1\right)}^2-{\left(\tan A\tan B\right)}^2} \\ =\frac{{\tan }^{2}A-{\tan }^{2}B}{1-{\tan }^{2}A{\tan }^{2}B} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$