If cos theta= cos alpha cos beta/1-sin alpha sin beta prove that sin t

1.	If cos theta= cos alpha cos beta/1-sin alpha sin beta prove that sin theta=±sin alpha+sin beta/1+sin alpha×sin beta
Answer» There's an error in the question as per my solution. You can check it. It should be : PROVE that sin theta = +/- sin alpha - sin BETA / 1 - sin alpha * sin beta. COS t = cos a cos B / 1 - sin a sin b sin²t = 1 - cos²t = 1 - (cos a cos b)² / (1 - sin a sin b)² = 1 - 2 sin a sin b + sin²a sin²b - cos²a cos²b / (1 - sin a sin b)² put 1 = sin²a + cos²a = sin²a + cos²a - 2 sin a sin b + sin²a sin²b - cos²a cos²b / (1 - sin a sin b)² = sin²a + cos²a( 1 - cos²b ) - 2 sin a sin b + sin²a sin²b / ( 1 - sin a sin b)² = sin²a + cos²a sin²b - 2 sin a sin b + sin²a sin²b / (1 - sin a sin b)² = sin²a + sin²b - 2 sin a sin b / (1 - sin a sin b)² = ( sin a - sin b )² / ( 1 - sin a sin b )² Hence sin t = +/- sin a - sin b / 1 - sin a sin b

If cos theta= cos alpha cos beta/1-sin alpha sin beta prove that sin theta=±sin alpha+sin beta/1+sin alpha×sin beta

Answer»

There's an error in the question as per my solution. You can check it.

It should be : PROVE that sin theta = +/- sin alpha - sin BETA / 1 - sin alpha * sin beta.

COS t = cos a cos B / 1 - sin a sin b

sin²t = 1 - cos²t

= 1 - (cos a cos b)² / (1 - sin a sin b)²

= 1 - 2 sin a sin b + sin²a sin²b - cos²a cos²b / (1 - sin a sin b)²

put 1 = sin²a + cos²a

= sin²a + cos²a - 2 sin a sin b + sin²a sin²b - cos²a cos²b / (1 - sin a sin b)²

= sin²a + cos²a( 1 - cos²b ) - 2 sin a sin b + sin²a sin²b / ( 1 - sin a sin b)²

= sin²a + cos²a sin²b - 2 sin a sin b + sin²a sin²b / (1 - sin a sin b)²

= sin²a + sin²b - 2 sin a sin b / (1 - sin a sin b)²

= ( sin a - sin b )² / ( 1 - sin a sin b )²

Hence sin t = +/- sin a - sin b / 1 - sin a sin b