Question

sin theta + cos theta / sin theta - cos theta + sin theta - cos theta/sin theta + cos theta = 2/2 sin^2 theta - 1.​

Answer 1

Answer:-

Sin¶ + Cos¶ / Sin¶ - Cos¶ +

Sin¶ - Cos¶ / Sin¶ + Cos¶ =

2/2 Sin²¶ - 1

Solving LHS we get,

( Sin¶ + Cos¶ )² + ( Sin¶ - Cos¶ )²

/ Sin²¶ - Cos²¶

= 2 / Sin²¶ - Cos²¶

Comparing LHS by RHS we get,

2 / Sin²¶ - Cos²¶ = 2/2 Sin²¶ - 1

2 / Sin²¶ - Cos²¶ = Sin²¶ - 1

2 / Sin²¶ - Cos²¶ = - Cos²¶

2 / Sin²¶ = 0

Sin²¶ = 2/0

Sin¶ = √2/0

¶ = Not defined

¶ = 90° ( ¶ is not defined when Sin¶ is 90° )

Note:- Please write question properly, it will be benificial for you only

Answer 2

Hope, it will help you....