Question

If $\tan \theta =\sqrt{3}-\sqrt{2}$, find ${\sin }^2\theta +{\sec }^2\theta -{cos}^2\theta$

Answer 1

$\tan \theta =\sqrt{3}-\sqrt{2}$

${\tan }^2\theta =3+2-2\sqrt{3}=5-2\sqrt{3}$

${\sin }^2\theta -{\sec }^2\theta -{\cos }^2\theta$

$\frac{2\tan \theta }{1+{\tan }^2\theta }+(1+{\tan }^2\theta )-\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$

$=\frac{2(\sqrt{3}-\sqrt{2})}{1+5-2\sqrt{3}}+1+5-2\sqrt{3}-\left(\frac{1-5+2\sqrt{3}}{1+5-2\sqrt{3}}\right)$

$=2\left(\frac{\sqrt{3}-\sqrt{2}}{2(3-\sqrt{3})}\right)+6-2\sqrt{3}-\left(\frac{-4+2\sqrt{3}}{6-2\sqrt{3}}\right)$

$=\frac{\sqrt{3}-\sqrt{2}}{3-\sqrt{3}}+6-2\sqrt{3}+\frac{2(2-\sqrt{3})}{2(3-\sqrt{3})}$

$=\frac{\sqrt{3}-\sqrt{2}+2-\sqrt{3}}{3-\sqrt{3}}+6-2\sqrt{3}$

$=\frac{2-\sqrt{2}}{3-\sqrt{3}}+6-2\sqrt{3}$

$=\frac{2-\sqrt{2}}{3-\sqrt{3}}+2(3-\sqrt{3})$

$=\frac{2-\sqrt{2}+2(9+3-2\sqrt{3})}{3-\sqrt{3}}$

$=\frac{2-\sqrt{2}+18+6-4\sqrt{3}}{3-\sqrt{3}}$

$=\frac{26-\sqrt{2}-4\sqrt{3}}{3-\sqrt{3}}$.