Question

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

Answer 1

Hi ,

I used A instead of theta .

LHS=1/( sinA + cosA) + 1/(sinA - cosA)

= (SinA-cosA +sinA+cosA)/(sinA+cosA)(sinA-cosA)

= 2sinA/ [ sin² A - cos² A)

= 2sinA/[-(cos² A- sin² A)]

= 2sinA / ( - cos2A)

= 2sinA / [ - ( 2cos²A- 1 ) ]

= 2sinA / ( 1 - 2cos²A)

= RHS

I hope this helps you.

:)