What is $tan^2theta/(1-sin^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2018-05-31T08:50:53

Please see the explanation below

We need

$cos^2x+sin^2x=1$

$tan^2x+1=sec^2x$

Therefore,

$tan^2x/(1-sin^2x)=(sec^2x-1)/(1-sin^2x)$

$=((secx+1)(secx-1))/((1+sinx)(1-sinx))$

What is tan^2theta/(1-sin^2thetatan^2theta/(1-sin^2theta in terms of non-exponential trigonometric functions?