Question

ABC is a righ angled triangle at b. given that ABC= tita side AB =2 BC=1 find sin2tita+tan2tita

Answer 1

In rt ∆ABC, AC = √(2×2+1×1) =√5
Then,
${ \sin}^{2} ( \beta ) + { \tan}^{2} ( \beta ) \\ = (\frac{2}{ \sqrt{5} })^{2} + ( \frac{2}{1})^{2} \\ = \frac{4}{5} + 4 \\ = \frac{4 + 20}{5} \\ = \frac{24}{5}$