How do you simplify #sin^2theta-cos^2theta+tan^2theta# to non-exponential trigonometric functions?

Question

7580 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-08T03:07:12

The given expression

#=sin^2theta-cos^2theta+tan^2theta#

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

Now recalling the identities

#(cos^2theta-sin^2theta)=cos2theta#

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

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

The given expression becomes

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

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

#=(1-2cos2theta-1/2(1+cos4theta))/(1+cos2theta)#

#=(1/2-2cos2theta-1/2cos4theta)/(1+cos2theta)#

#=(1-4cos2theta-cos4theta)/(2(1+cos2theta))#