Question #c20a4

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Answer 1 · 2017-04-04T23:31:32

I tried this:

Have a look:
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Answer 2 · 2017-04-04T23:32:25

$(tanx+cotx)/(tanx-cotx)=sec^2x/(tan^2x-1)$

Multiply the left-hand side by the conjugate of the denominator:

$(tanx+cotx)/(tanx-cotx)=(tanx+cotx)/(tanx-cotx)*(tanx+cotx)/(tanx+cotx)$

$=(tan^2x+2tanxcotx+cot^2x)/(tan^2x-cot^2x)$

Rewriting $cotx$ :

$=(tan^2x+2tanx(1/tanx)+1/tan^2x)/(tan^2x-1/tan^2x)$

Multiplying through by $tan^2x$ :

$=(tan^4x+2tan^2x+1)/(tan^4x-1)$

Factoring:

$=(tan^2x+1)^2/((tan^2x+1)(tan^2x-1))=(tan^2x+1)/(tan^2x-1)$

The numerator is a form of the Pythagorean identity:

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

This is the right-hand side of the original equation, so the identity has been verified.