Question

Write down trigonometric ratios of angle a in terms of tan a?

Answer 1

$se{c}^2=1+ta{n}^{2}a\Rightarrow seca=\sqrt{1+ta{n}^2}-\left(1\right)\frac{1}{cosa}=\sqrt{1+ta{n}^{2}a}\Rightarrow cosa=\frac{1}{\sqrt{1+ta{n}^{2}a}}-\left(2\right)sina=\sqrt{1-co{s}^{2}a}=\sqrt{1-\frac{1}{1+ta{n}^{2}a}}=\sqrt{\frac{ta{n}^{2}a}{1+ta{n}^{2}a}}=\frac{tana}{\sqrt{1+ta{n}^{2}a}}-\left(3\right)$
$coseca=\frac{1}{sina}=\frac{\sqrt{1+ta{n}^{2}a}}{tana}-\left(4\right)tana=tana-\left(5\right)cota=\frac{1}{tana}-\left(6\right)$