How do you evaluate ${sec}^{2} (18) - {tan}^{2} (18)$ $sec^2(18) - tan^2(18)$ ?

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Answer 1 · 2018-05-21T03:19:33

${sec}^{2} (18) - {tan}^{2} (18)$ $sec^2(18) - tan^2(18)$

$= \frac{1}{{cos}^{2} (18)} - \frac{{sin}^{2} (18)}{{cos}^{2} (18)}$ $= 1/cos^2(18) - sin^2(18)/cos^2(18)$

$= \frac{1 - {sin}^{2} (18)}{{cos}^{2} (18)}$ $= (1-sin^2(18))/cos^2(18)$

$= \frac{{cos}^{2} (18)}{{cos}^{2} (18)}$ $= cos^2(18)/cos^2(18)$

$= 1$ $= 1$

How do you evaluate sec^2(18) - tan^2(18)sec2(18)−tan2(18)sec^2(18) - tan^2(18)?