Question

If cosA/cosB=n and sinA/sinB=m then (m^2-n^2)=?​

Answer 1

Answer:

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

Answer 2

given data

cosA/cosB = n ----(1

sinA/sinB = m -----(2

on squring both the equation

m^2 = (sin^2A/sin^2B )

n^2 = (cos^2A/cos^2B)

Now

m^2 - n^2

sin^2A/sin^2B - cos^2A/cos^2B

(sin^2Acos^2B - cos^2ASin^2B)/sin^2Acos^2A

4sin(A - B)sin(A + B)/sin^2(2A)