Use the trigonometric substitution u = a tan θ, where 0

Question

Use the trigonometric substitution u = a tan θ, where 0 < θ < π/2 and to simplify the expression  . ​
A)​- a sec θ 
B)​a sec θ 
C)​a tan θ 
D)​a sin θ 
E)​a csc θ

Quizplus · Accepted Answer

Using the trigonometric substitution u = a tan θ, we need to convert the given expression to involve θ. We can do this by using the identity:

$$\sec^2	heta = 1+	an^2	heta$$

Multiplying and dividing the expression by $\sqrt{a^2 + x^2}$, we get:

$$\frac{\sqrt{a^2+x^2}}{x} = \frac{\sqrt{a^2+x^2}}{a	an	heta}$$

$$= \frac{\sqrt{a^2+x^2}}{a} \cdot \frac{1}{	an	heta}$$

$$= \frac{\sqrt{a^2+x^2}}{a} \cdot \frac{\cos	heta}{\sin	heta}$$

Now, using the identity $	an	heta = \frac{\sin	heta}{\cos	heta}$, we get:

$$= \frac{\sqrt{a^2+x^2}}{a} \cdot \frac{\cos	heta}{\frac{\sin	heta}{\cos	heta}}$$

$$= \frac{\sqrt{a^2+x^2}}{a} \cdot \frac{\cos^2	heta}{\sin	heta}$$

Recall that $\cos^2	heta = 1-\sin^2	heta$. Using the identity $\sin^2	heta = \frac{a^2}{a^2+x^2}$ (which follows from $u = a	an	heta$), we get:

$$= \frac{\sqrt{a^2+x^2}}{a} \cdot \frac{1-\frac{a^2}{a^2+x^2}}{\sqrt{\frac{a^2}{a^2+x^2}}}$$

$$= \frac{\sqrt{a^2+x^2}}{a} \cdot \frac{\sqrt{a^2+x^2}-a}{a}$$

Simplifying, we get:

$$\frac{\sqrt{a^2+x^2}}{x} = \frac{\sqrt{a^2+x^2}-a}{a\sin	heta} = \frac{1}{a\sin	heta} - \frac{1}{\cos	heta}$$

Substituting back $\sin	heta = \frac{u}{a}$ and $\cos	heta = \frac{1}{\sqrt{1+	an^2	heta}} = \frac{1}{\sqrt{1+\frac{u^2}{a^2}}}$, we get:

$$\frac{\sqrt{a^2+x^2}}{x} = \frac{1}{au} - \frac{a}{\sqrt{a^2+u^2}}$$

Thus, the answer is B) a sec θ.

Use the Trigonometric Substitution U = a Tan θ, Where