Question

Cosa/cosb=n and sina/sinb=m then (m^2-n^2)sin^b=

Answer 1

According to question
m = Sin a/Sin b
n = Cos a/Cos b

thus putting these in 
(m^2 - n^2)Sin^2b = (Sin^2 a/Sin^2 b - Cos^2 a/Cos^2 b)Sin^2 b

taking LCM
                              = [(Sin^2 a Cos^2 b - Cos^2 a Sin^2 b)/Sin^2 b Cos^2 b]
                                       * Sin^2 b
                              = (Sin^2 a Cos^2 b - Cos^2 a Sin^2 b)/ Cos^2 b
                              = Sin^2 a - Cos^2 a Sin^2 b/Cos^2 b
                             = Sin^2 a -  Cos^2 a tan^2 b