Question

sina-sinb/cos a+cos b + cos a-cos b/sin a +sin b

Answer 1

$$\frac {\sin a - \sin b}{\cos a + \cos b} + \frac {\cos a - \cos b}{\sin a + \sin b} = \frac {(\sin a - \sin b) (\sin a + \sin b) + (\cos a - \cos b) (\cos a + \cos b)}{(\cos a + \cos b) (\sin a + \sin b)}$$

We know that $(a - b)(a + b) = a^2 - b^2$ (2)

So, using formula (2):

$$(\sin a - \sin b) (\sin a + \sin b) = \sin^ {2} a - \sin^ {2} b$$ $$(\cos a - \cos b) (\cos a + \cos b) = \cos^ {2} a - \cos^ {2} b$$

(3),(4) $\rightarrow$ (1)

$$\begin{array}{l} \frac {(\sin a - \sin b) (\sin a + \sin b) + (\cos a - \cos b) (\cos a + \cos b)}{(\cos a + \cos b) (\sin a + \sin b)} = \frac {\sin^ {2} a - \sin^ {2} b + \cos^ {2} a - \cos^ {2} b}{(\cos a + \cos b) (\sin a + \sin b)} = \\ = \frac {\sin^ {2} a + \cos^ {2} a - \sin^ {2} b - \cos^ {2} b}{(\cos a + \cos b) (\sin a + \sin b)} = \frac {(\sin^ {2} a + \cos^ {2} a) - (\sin^ {2} b + \cos^ {2} b)}{(\cos a + \cos b) (\sin a + \sin b)} \\ \end{array}$$

We knew: $\sin^2 x + \cos^2 x = 1$ (6)

(6) $\rightarrow$ (5)

$$\begin{array}{l} \frac {(\sin^ {2} a + \cos^ {2} a) - (\sin^ {2} b + \cos^ {2} b)}{(\cos a + \cos b) (\sin a + \sin b)} = \frac {1 - 1}{(\cos a + \cos b) (\sin a + \sin b)} \\ = \frac {0}{(\cos a + \cos b) (\sin a + \sin b)} = 0 \\ \end{array}$$