verify this trigonometric identity

Question

cos 'a' - cos 'b' / sin 'a' + sin 'b'
+
sin 'a' - sin 'b' / cos 'a' + cos 'b'
= 0

please help me with this! i don't know how to do it!

Answer 1

I am sure that you meant:
(cos 'a' - cos 'b') / (sin 'a' + sin 'b')
+
(sin 'a' - sin 'b') / (cos 'a' + cos 'b')
= 0

the LCD is (sina + sinb)(cosa + cosb)

LS = ( (cosa - cosb)(cosa + cosb) + sina - sinb)(sina - sinb) ) / ( (sina + sinb)(cosa + cosb) )
= (cos^2 a - cos^2 b + sin^2 a - sin^2 b) / ( (sina + sinb)(cosa + cosb) )
= ( cos^2 a + sin^2 a - (cos^2 ba + sin^2 b) / ( (sina + sinb)(cosa + cosb) )
= (1-1)/( (sina + sinb)(cosa + cosb) )
= 0/( (sina + sinb)(cosa + cosb) )
= 0
= RS

Answer 2

For clutter reduction, let's let

x = (a+b)/2
y = (a-b)/2

using the sum-to-product formulas,

cosa-cosb = -2 sinx siny
sina+sinb = 2 sinx cosy
sina-sinb = 2 cosx siny
cosa+cosb = 2 cosx cosy

Now we just have

-tany + tany = 0
seems legit to me.