Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of $\theta $, where $0

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of $\theta $, where $0 < x < \frac { \pi } { 2 }$ . $\sqrt { x ^ { 2 } + 25 } , x = 5 \tan \theta$  
A)- 5 sec$\theta $
B)5 tan $\theta $
C)25 sec $\theta $
D)25 tan $\theta $
E)5 sec$\theta $

Quizplus · Accepted Answer

Substituting $x = 5\tan\theta$ into the expression $\sqrt{x^2 + 25}$ gives $\sqrt{(5\tan\theta)^2 + 25} = \sqrt{25\tan^2\theta + 25} = \sqrt{25(\tan^2\theta + 1)}$. Using the trigonometric identity $\tan^2\theta + 1 = \sec^2\theta$, we get $\sqrt{25\sec^2\theta} = 5\sec\theta$.

Use the Trigonometric Substitution to Rewrite the Algebraic Expression as a Trigonometric