How do you express #f(theta)=-sin^2(theta)+cot^2(theta)-2tan^2theta# in terms of non-exponential trigonometric functions?

Question

324 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-19T10:37:03

See the explanation below

Reminder :

#sin^2theta=(1-cos2theta)/2#

#cos^2theta=(1+cos2theta)/2#

#cottheta=costheta/sintheta#

#tantheta=sintheta/costheta#

Therefore,

#f(theta)=-sin^2theta+cot^2theta-2tan^2theta#

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

#=-(1-cos2theta)/2+((1+cos2theta)/2)/((1-cos2theta)/2)-2((1-cos2theta)/2)/((1+cos2theta)/2)#

#=-(1-cos2theta)/2+((1+cos2theta))/((1-cos2theta))-2((1-cos2theta))/((1+cos2theta))#