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 $0 < x < \frac { \pi } { 2 }$ . $\sqrt { 25 x ^ { 2 } + 36 } , 5 x = 6 \tan \theta$  
A)25 tan$\theta $
B)5 tan $\theta $
C)25 sec$\theta $
D)6 sec $\theta $
E)- 5 sec $\theta $

Quizplus · Accepted Answer

Using the substitution $5x = 6\tan\theta$, we have $x = \frac{6}{5}\tan\theta$. Substitute into the expression: $\sqrt{25x^2 + 36} = \sqrt{25(\frac{6}{5}\tan\theta)^2 + 36} = \sqrt{36\tan^2\theta + 36} = \sqrt{36(\tan^2\theta + 1)} = \sqrt{36\sec^2\theta} = 6\sec\theta$.

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