Question

$ABC$ is a right-angled triangle, right-angled at $B$/. Given that $\angle ACB=\theta$, side $AB=2$ units and side $BC=1unit$, find the value of ${\sin }^2\theta +{\tan }^2\theta$.

Answer 1

$\sin \theta =\frac{AC}{A{B}^2};\tan \theta =\frac{AC}{BC}$

$\left[\begin{array}{c}\because AB=\sqrt{5}\\ AC=2\\ BC=1\end{array}\right]$

In $△ABC,\angle B={90}^o$ By Pythagoras theorem,

$\therefore \sin \theta =\frac{AC}{AB}$

$\sin \theta =\frac{2}{\sqrt{5}}$

$\tan \theta =\frac{2}{1}$

$A{B}^2=A{C}^2+B{C}^2$

$A{B}^2=2^2+1^2$

$AB=\sqrt{5}$

$\therefore {\sin }^2\theta +{\tan }^2\theta ={\left(\frac{2}{\sqrt{5}}\right)}^2+{\left(\frac{2}{1}\right)}^2$

$=\frac{4}{5}+\frac{4}{2}=\frac{4+4\times 5}{5}=\frac{24}{5}$