Question

tanθ + sin θ / tan θ - sin θ= secθ +1= sec θ-1,Prove it by using trigonometric identities.

Answer 1

I think question is ---> (tanθ + sinθ)/ (tanθ - sinθ) = (secθ +1)/(sec θ-1)

LHS = (tanθ + sinθ)/(tanθ - sinθ)

= (sinθ/cosθ + sinθ)/(sinθ/cosθ - sinθ)

= (sinθ + sinθ.cosθ)/(sinθ - sinθ.cosθ)

= sinθ(1 + cosθ)/sinθ(1 - cosθ)

= (1 + cosθ)/(1 - cosθ)

= (1 + 1/secθ)/(1 - 1/secθ)

= (secθ + 1)/(secθ - 1) = RHS

Answer 2

Hi ,

Here I am using ' A ' instead of theta.

LHS = ( tan A + sinA )/( Tan A - SinA )

divide denominator and numerator with

sin A , we get

=(tanA/sinA+sinA/sinA)/(tanA/sinA-sinA/sinA)

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

= ( secA + 1 )/( secA - 1 )

= RHS

I hope this helps you.

: )