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

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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

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

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

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

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

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

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