How do you prove $tan^2(x) / (sec(x) - 1) = (sec(x) + 1)$ ?

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Answer 1 · 2015-12-13T17:19:22

The Pythagorean identity

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

can be divided by $cos^2x$ to see that

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

Then, $1$ can be subtracted from either side to see that

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

So, the identity on the left can be rewritten as

$(sec^2x-1)/(secx-1)$

Now, see that $sec^2x-1$ is a difference of squares which can be factored into

$(secx+1)(secx-1)$

Substitute this into the expression to get

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

Which simplifies to be

$(secx+1)$

How do you prove tan^2(x) / (sec(x) - 1) = (sec(x) + 1) tan^2(x) / (sec(x) - 1) = (sec(x) + 1)?