Question

In a right angle triangle ABC right angle at B , $\angle ABC=\theta$ , AB = 2cm and BC = 1cm . Find the value of $si{n}^2\theta +ta{n}^2\theta$

Answer 1

We have,

$AB=2cm$ and $BC=1cm$

Using Pythagoras theorem in $\Delta$CAB, we have

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

$\Rightarrow A{C}^2=2^2+1=5$

$\Rightarrow AC=\sqrt{5}$

Thus, Base$=BC=1cm,$ Perpendicular $=AB=2cm$

and, Hypotenuse $=AC=\sqrt{5}cm$

$sin\theta =\frac{AB}{AC}=\frac{2}{\sqrt{5}},$ and tan $\theta =\frac{AB}{BC}=\frac{2}{1}=2$

Now, $si{n}^2\theta +ta{n}^2\theta =(sin\theta {)}^2+(tan\theta {)}^2={\left(\frac{2}{\sqrt{5}}\right)}^2+(2{)}^2=\frac{4}{5}+4=\frac{24}{5}$