Question

Given the trig substitution $x = \frac{3}{2}sec(\theta)$, evaluate the expression $tan(\theta)$ in terms of $x$. $\tan(\theta) = \frac{2x}{\sqrt{9-4x^2}}$ $\tan(\theta) = \frac{\sqrt{4x^2-9}}{3}$ $\tan(\theta) = \frac{\sqrt{4x^2-9}}{2x}$ $\tan(\theta) = \frac{\sqrt{9+4x^2}}{3}$

          Given the trig substitution $x = \frac{3}{2}sec(\theta)$, evaluate the expression $tan(\theta)$ in terms of $x$.
$\tan(\theta) = \frac{2x}{\sqrt{9-4x^2}}$
$\tan(\theta) = \frac{\sqrt{4x^2-9}}{3}$
$\tan(\theta) = \frac{\sqrt{4x^2-9}}{2x}$
$\tan(\theta) = \frac{\sqrt{9+4x^2}}{3}$

Given the trig substitution x = (3)/(2)sec(θ), evaluate the expression tan(θ) in terms of x.
tan(θ) = (2x)/(√(9-4x^2))
tan(θ) = (√(4x^2-9))/(3)
tan(θ) = (√(4x^2-9))/(2x)
tan(θ) = (√(9+4x^2))/(3)

Added by Amy C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Linh Vu

05/14/2020

Step 1

$x = \frac{3}{2}sec(\theta)$ $sec(\theta) = \frac{2x}{3}$ Show more…

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Given the trig substitution $x = \frac{3}{2}sec(\theta)$, evaluate the expression $tan(\theta)$ in terms of $x$. tan(0) = $\frac{2x}{\sqrt{9-4x^2}}$ tan(0) = $\frac{\sqrt{4x^2-9}}{3}$ tan(0) = $\frac{\sqrt{4x^2-9}}{2x}$ tan(0) = $\frac{\sqrt{9+4x^2}}{3}$