Question

Show that: $\frac{(\sin A-\sin B)}{(\cos A+\cos B)}+\frac{(\cos A-\cos B)}{(\sin A+\sin B)}=0$

Answer 1

Consider,

LHS: $\frac{(\sin A-\sin B)}{(\cos A+\cos B)}+\frac{(\cos A-\cos B)}{(\sin A+\sin B)}$

$=\frac{(\sin A-\sin B)(\sin A+\sin B)+(\cos A-\cos B)(\cos A+\cos B)}{(\cos A+\cos B)(\sin A+\sin B)}$

$=\frac{{\sin }^{2}A+\sin A\sin B-\sin B\sin A-{\sin }^{2}B+{\cos }^{2}A+\cos A\cos B-\cos A\cos B-{\cos }^{2}B}{(\cos A+\cos B)(\sin A+\sin B)}$

$=\frac{({\sin }^{2}A+{\cos }^{2}A)-({\sin }^{2}B+{\sin }^{2}A)}{(\cos A+\cos B)(\sin A+\sin B)}$

$=\frac{0}{(\cos A+\cos B)(\sin A+\sin B)}$

$=0$

$=$ RHS

Hence, proved.