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

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where $0 < x < \frac { \pi } { 2 }$ . $\sqrt { 25 x ^ { 2 } + 36 } , 5 x = 6 \tan \theta$ ​
A)25 tan θ
B)5 tan θ
C)25 sec θ
D)6 sec θ
E)- 5 sec θ

Quizplus · Accepted Answer

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

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