Let x=asin⁡θ in a2−x2. Then find cos⁡θ tan⁡θ

$\sqrt{a^2-s^2{\sin }^2(\theta )}=$
Factor out common term $a^2$
$=\sqrt{a^2(1-{\sin }^2(\theta ))}$
Apply radical rule $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$, assuming $\frac{a\ge 0}{b\ge 0}$
$=\sqrt{a^2}\sqrt{-{\sin }^2(\theta )+1}$
Apply radical rule $\sqrt[n]{a^n}=a$, assuming $a\ge 0$ Use the following identity: ${\cos }^2(x)+{\sin }^2(x)=1$
Therefore $1-{\sin }^2(x)={\cos }^2(x)$
$=a\sqrt{{\cos }^2(\theta )}=a\cos (\theta )$
Therefore $\sqrt{a^2-a^2{\sin }^2(\theta )}=a\cos (\theta )$
$⟹\cos \theta =\sqrt{a^2-a^2}\frac{{\sin }^2(\theta )}{a}$
$⟹\frac{a\sin (\theta )}{\sqrt{a^2-a^2{\sin }^2(\theta )}}=\tan (\theta )$
$\sqrt{a^2-a^2{\sin }^2(\theta )}=a\cos (\theta )$