Simplify $\frac{{tan}^{2} x}{sec x + 1}$ $tan^2x/(secx+1)$ ?

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Answer 1 · 2017-03-21T01:05:23

Please see below.

We do it using wellKnown identity ${sec}^{2} x = {tan}^{2} x + 1$ $sec^2x=tan^2x+1$ or ${sec}^{2} x - 1 = {tan}^{2} x$ $sec^2x-1=tan^2x$ . Using this

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

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

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

= $sec x - 1$ $secx-1$ .

Simplify tan^2x/(secx+1)tan2xsecx+1tan^2x/(secx+1)?