Question

(1-sin theta/1+sin theta) =(sec theta-tan theta) ^2 pls solve​

Answer 1

Answer:

RHS=(secθ -tanθ)

LHS =1+sinθ1−sinθ 

Multiply and divide by (1−sinθ)

 = 1 - sinθ/1+sinθ × 1 - sinθ/1 - sinθ

=(1−sinθ)^2/cos^2 θ

= (1 - sinθ/cosθ)^2

 = (secθ−tanθ)^2

 LHS=RHS 

Hence proved

Answer 2

Refer to the attachment...