Question 1196469
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If {{{x = sec^2(theta)}}} then {{{1/x = 1/(sec^2(theta)) = cos^2(theta)}}}


Check out this list of trig identities
<a href = "https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf</a>
Look under the "pythagorean identities" subsection, on page 2, to find the identity {{{1 + cot^2(theta) = csc^2(theta)}}}
This rearranges to {{{cot^2(theta) = csc^2(theta)-1}}}


So we could say:
{{{y+2 = cot^2(theta)}}}


{{{y+2 = csc^2(theta)-1}}}


{{{y = csc^2(theta)-1-2}}}


{{{y = csc^2(theta)-3}}}


{{{y = 1/(sin^2(theta))-3}}}


{{{y = 1/(1-cos^2(theta))-3}}}


{{{y = 1/(1-1/x)-3}}}


{{{y = 1/(x/x-1/x)-3}}}


{{{y = 1/((x-1)/x)-3}}}


{{{y = x/(x-1) - 3}}}


We could optionally continue like so
{{{y = x/(x-1) - 3}}}


{{{y = x/(x-1) - 3(x-1)/(x-1)}}}


{{{y = (x - 3(x-1))/(x-1)}}}


{{{y = (x - 3x+3)/(x-1)}}}


{{{y = (-2x+3)/(x-1)}}}
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