How do you simplify the expression $\frac{{tan}^{2} x}{sec x - 1}$ $tan^2x/(secx-1)$ ?

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Answer 1 · 2016-09-26T10:36:56

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

As ${sec}^{2} x = 1 + {tan}^{2} x$ $sec^2x=1+tan^2x$ , ${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x=sec^2x-1$

Hence, $\frac{{tan}^{2} x}{sec x - 1}$ $tan^2x/(secx-1)$

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

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

How do you simplify the expression tan^2x/(secx-1)tan2xsecx−1tan^2x/(secx-1)?