Question

if x=4cotθ, use trigonometric substitution to write √(16+x^2) as a trigonometric function of θ, where 0<θ 12 years ago

Answer 1

\[\large \color{royalblue}{x=4\cot \theta}\]

\[\large \sqrt{16+\color{royalblue}{x}^2} \qquad=\qquad \sqrt{16+\color{royalblue}{16\cot^2\theta}} \qquad=\qquad \sqrt{16(\color{orangered}{1+\cot^2\theta})}\]

Recalling one of our square identities,\[\large \color{orangered}{1+\cot^2\theta=\csc^2\theta}\]