Question

If $sin\theta =\frac{3}{5},$ find the value of $(tan\theta +sec\theta {)}^2$

Answer 1

$\therefore cos\theta =\sqrt{1-si{n}^2\theta }=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
$\therefore tan\theta =\frac{sin\theta }{cos\theta }=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$ and $sec\theta =\frac{1}{cos\theta }\Rightarrow sec\theta =\frac{5}{4}$
$\therefore (tan\theta +sec\theta {)}^2={(\frac{3}{4}+\frac{5}{4})}^2={\left(\frac{8}{4}\right)}^2=4$