What is #sectheta+cos^2theta -costheta# in terms of #sintheta#?

Question

1687 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-22T07:06:20

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

#sec(theta) + cos^2(theta) - cos(theta)#

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

#=1/cos(theta) + cos^2(theta) - cos(theta)#

#=1/cos(theta) + (cos^2(theta)cos(theta))/cos(theta) - (cos(theta)cos(theta))/cos(theta)#

#= (1+cos^3(theta)-cos^2(theta))/cos(theta)#

#=(1-cos^2(theta) + cos(theta)cos^2(theta))/cos(theta)#

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

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)#

#=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)#