Question

$\tan \: a = n \: \tan \: b \: and \: \sin \: a = m \: sin \: b \\ prove \: that \: {cos}^{2} a = \frac{ {m}^{2} - 1 }{n^{2} - 1 } .$
help guyz...urgently

Answer 1

Given tan a = n tan b ----------- (1)

Given sin a = m sin b ----------- (2).

Equation (1) can be written as 

tan a/tan b = n

(sin a/cos a) * (cos b * sin b) = n

(sin a * cos b/(cos a * sin b) = n   -------------------- (3)

Substitute (3) in (2), we get

(m sin b * cos b)/(cos a * sin b) = n

m cos b/cos a = n

m cos b = n cos a

Squaring on both sides, we get

m^2 cos^2 b = n^2 cos^2 a

m^2(1 - sin^2 b) = n^2 cos^2 a

m^2(1 - sin^2 a/m) = n^2 cos ^2 a  (From (2))

m^2 - sin^2 a = n^2 cos^2 a

m^2 - 1 + cos^2 a = n^2 cos^2 a

m^2 - 1 = n^2 cos^2 a - cos^2 a

m^2 - 1 = cos^2 a(n^2 - 1)

cos^2 a = m^2 - 1/n^2 - 1.


Hope this helps!

Answer 2

this is the half solution but other part is not loading so plz check in my question I uploaded the other part.