Question 1206148
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{{{(tan(theta))/(1 - cos^2(theta))}}}


= {{{(tan(theta))/(sin^2(theta))}}} Use the pythagorean trig identity


= {{{matrix(1,3,tan(theta),"divide",sin^2(theta))}}}


= {{{matrix(1,3,sin(theta)/cos(theta),"divide",sin^2(theta))}}}


= {{{(sin(theta)/cos(theta))*(1/(sin^2(theta)))}}}


= {{{(1/cos(theta))*(1/sin(theta))}}}


= {{{sec(theta)*csc(theta)}}}


Therefore,
{{{(tan(theta))/(1 - cos^2(theta)) = sec(theta)*csc(theta)}}}
is an identity.


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Alternatively,


{{{(1/cos(theta))*(1/sin(theta))}}}


= {{{1/(cos(theta)*sin(theta)))}}}


= {{{1/(0.5sin(2theta)))}}} Use a variation of sin(2theta) = 2*sin(theta)*cos(theta)


= {{{2/(sin(2theta))}}}


= {{{2csc(2theta)}}}


So we could also say that
{{{(tan(theta))/(1 - cos^2(theta)) = 2csc(2theta)}}}
is an identity.


You can use a graphing calculator to verify each identity. 
Desmos is one good choice.
<a href="https://www.desmos.com/calculator/jwy2tauhgg">https://www.desmos.com/calculator/jwy2tauhgg</a>
Notice how one curve overlaps the other perfectly. Click each curve on/off to help show the overlap (the curve should blink different colors).
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