What is sectheta+cos^2theta -costhetasecθ+cos2θcosθ in terms of sinthetasinθ?

1 Answer
Jan 22, 2016

=sin^2(theta)/sqrt(1-sin^2(theta)) + 1- sin^2(theta)=sin2(θ)1sin2(θ)+1sin2(θ)

Explanation:

sec(theta) + cos^2(theta) - cos(theta)sec(θ)+cos2(θ)cos(θ)

To simplify in terms of sin(theta)sin(θ) let us write sec(theta)sec(θ) as 1/cos(theta)1cos(θ)

=1/cos(theta) + cos^2(theta) - cos(theta)=1cos(θ)+cos2(θ)cos(θ)

=1/cos(theta) + (cos^2(theta)cos(theta))/cos(theta) - (cos(theta)cos(theta))/cos(theta)=1cos(θ)+cos2(θ)cos(θ)cos(θ)cos(θ)cos(θ)cos(θ)

= (1+cos^3(theta)-cos^2(theta))/cos(theta)=1+cos3(θ)cos2(θ)cos(θ)

=(1-cos^2(theta) + cos(theta)cos^2(theta))/cos(theta)=1cos2(θ)+cos(θ)cos2(θ)cos(θ)

=(sin^2(theta)+sqrt(1-sin^2(theta))(1-sin^2(theta)))/sqrt(1-sin^2(theta)=sin2(θ)+1sin2(θ)(1sin2(θ))1sin2(θ)

If you need it can be simplified further as

=sin^2(theta)/sqrt(1-sin^2(theta)) + (sqrt(1-sin^2(theta))(1-sin^2(theta)))/sqrt(1-sin^2(theta)=sin2(θ)1sin2(θ)+1sin2(θ)(1sin2(θ))1sin2(θ)

=sin^2(theta)/sqrt(1-sin^2(theta)) + (cancel(sqrt(1-sin^2(theta)))(1-sin^2(theta)))/cancel(sqrt(1-sin^2(theta))

=sin^2(theta)/sqrt(1-sin^2(theta)) + 1- sin^2(theta)