Question

Trigonometric Substitution Make the indicated trigonometric substitution in the given algebraic expression and simplify. Assume that $\mathbf{0}\mathbf{\text{<}}\mathit{\theta }\mathbf{\text{<}}\frac{\mathbf{\pi }}{\mathbf{2}}$

$\frac{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{25}}}{\mathbf{x}}\mathbf{,}\mathit{x}\mathbf{=}\mathbf{5}\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }$

Short answer

The simplified expression is$\frac{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{25}}}{\mathbf{x}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

Step-by-step solution

Step 1. Given information

An expression $\frac{\sqrt{x^2-25}}{x}$ and value for substitution as$x=5\sec \theta$

Step 2. Concept used

To begin, change the value of$x$ in the equation in question. Then, using trigonometric identity, reduce it to its most basic form. The right answer is in its most basic form.

Step 3. Calculation

Now, substitute the value in$\frac{\sqrt{x^2-25}}{x}$ and simplify:

$\begin{array}{l}\frac{\sqrt{x^2-25}}{x}=\frac{\sqrt{{\left(5\sec \theta \right)}^2-25}}{5\sec \theta }\\ =\frac{\sqrt{25{\sec }^2\theta -25}}{5\sec \theta }\\ =\frac{\sqrt{25\left({\sec }^2\theta -1\right)}}{5\sec \theta }\\ \end{array}$

Now identities,

$\tan \theta =\frac{\sin \theta }{\cos \theta },\sec \theta =\frac{1}{\cos \theta }$

$\begin{array}{l}=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}\\ =\sin \theta \end{array}$

After simplification, we get-

$\sin \theta$