Question

PROVE THAT. secθ (1- sinθ) (secθ +tanθ) =1​

Answer 1

[sec(θ)+tan(θ) -1]/[tan(θ) -sec(θ )+1]=cos(θ)/[1-sin(θ)]

L.H.S=[sec(θ +tan(θ) -1]/[tan(θ) -sec(θ )+1]

        =[1 +sin(θ) -cos(θ)][sin(θ)-1 +cos(θ)]

        =[sin(θ) +1-[-2sin^2(θ/2)+1]]/[sin(θ) +1-2sin^2(θ/2)-1]

        =[2sin(θ/2)cos(θ/2)+2sin^2(θ/2)]/[2sin(θ/2).cos(θ/2)-2sin^2(θ/2)]

        = 2sin(θ/2)[sin(θ/2)+cos(θ/2)]/[2sin(θ/2)[cos(θ/2)-sin(θ/2)]    

       =[sin(θ/2)+cos(θ/2)]/[cos(θ/2)-sin(θ/2)]

        =[sin(θ/2)+cos(θ)]*[cos(θ/2)-sin(θ/2)]/[sin(θ/2) cos(θ/2)]^2

        =[cos^2(θ/2) - sin^2(θ/2)]/[sin^2(θ/2) +cos^2(θ/2) -2sin(θ/2).cos(θ/2)]      

        accordingly{[x+y]^2 = x^2 + y^2 +2xy} and sin^2(θ) +cos^2(θ) = 1

         =cos(θ)/[1 -2sin(θ/2)cos(θ/2)]

        =cos(θ)[1 -sin(θ)]//

         = 1

hence prov e