How do you prove $\frac{{tan}^{2} x}{sec x + 1} + 1 = sec x$ $tan^2x/(Secx+1)+1 = secx$ ?

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How do you prove $\frac{{tan}^{2} x}{sec x + 1} + 1 = sec x$ $tan^2x/(Secx+1)+1 = secx$ ?

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Answer 1 · 2015-04-15T23:45:31

Tiago Hands

Apr 15, 2015

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Answer 2 · 2016-10-20T12:12:00

For reasons explained in the video below, it turns out that:

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

Therefore:

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

Now, due to the FOIL rule (first, outer, inner, last)...

${sec}^{2} x - 1 = (sec x + 1) (sec x - 1)$ $sec^2x-1=(secx+1)(secx-1)$

All of the information above combined ultimately means that...

$L H S = \frac{{tan}^{2} x}{sec x + 1} + 1$ $LHS=(tan^2x)/(secx+1)+1$

$= \frac{{sec}^{2} x - 1}{sec x + 1} + 1$ $=(sec^2x-1)/(secx+1)+1$

$= \frac{(sec x + 1) (sec x - 1)}{sec x + 1} + 1$ $=((secx+1)(secx-1))/(secx+1)+1$

*You can now get rid of (secx+1) at the top and bottom of the fraction. When the numerator and denominator of a fraction are both the same, providing they aren't both zeros, what you get is 1.

$= sec x - 1 + 1$ $=secx-1+1$

$= sec x$ $=secx$

$= R H S$ $=RHS$

And here is your proof.

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How do you prove $\frac{{tan}^{2} x}{sec x + 1} + 1 = sec x$ $tan^2x/(Secx+1)+1 = secx$ ?

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