Solved on Jan 28, 2024

Find the secant squared of the inverse cotangent of $x$ . Select the correct answer.

STEP 1

Assumptions
1. We are looking to find the value of $\sec^2(\cot^{-1} x)$ .
2. $\cot^{-1} x$ represents the inverse cotangent function, which gives us an angle whose cotangent is $x$ .
3. The relationship between the trigonometric functions cotangent and secant will be used.
4. We will use the identity $\sec^2 \theta = 1 + \tan^2 \theta$ .

STEP 2

First, let's denote the angle whose cotangent is $x$ as $\theta$ . Therefore, we have:
$\cot \theta = x$

STEP 3

We know that the cotangent of an angle in a right triangle is the adjacent side over the opposite side. Let's represent the adjacent side as $x$ and the opposite side as $1$ (since $\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = x$ ).

STEP 4

Using the Pythagorean theorem, we can find the hypotenuse $h$ of the right triangle with sides $x$ and $1$ .
$h^2 = x^2 + 1^2$

STEP 5

Solve for the hypotenuse $h$ .
$h = \sqrt{x^2 + 1}$

STEP 6

Now, we can express the secant of $\theta$ in terms of $x$ and $h$ . Secant is the reciprocal of cosine, and cosine is the adjacent side over the hypotenuse. Therefore:
$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$

STEP 7

Substitute the value of $h$ into the expression for $\sec \theta$ .
$\sec \theta = \frac{\sqrt{x^2 + 1}}{x}$

STEP 8

Square the expression to find $\sec^2 \theta$ .
$\sec^2 \theta = \left(\frac{\sqrt{x^2 + 1}}{x}\right)^2$

STEP 9

Simplify the squared expression.
$\sec^2 \theta = \frac{x^2 + 1}{x^2}$

STEP 10

Now, we replace $\theta$ with $\cot^{-1} x$ to find $\sec^2(\cot^{-1} x)$ .
$\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}$

STEP 11

We can see that the expression matches one of the given options.
The solution is:
$\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}$
Therefore, the correct answer is $\frac{1+x^2}{x^2}$ .

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Find the secant squared of the inverse cotangent of $x$ . Select the correct answer.

STEP 1

STEP 2

First, let's denote the angle whose cotangent is $x$ as $\theta$ . Therefore, we have:
$\cot \theta = x$

STEP 3

We know that the cotangent of an angle in a right triangle is the adjacent side over the opposite side. Let's represent the adjacent side as $x$ and the opposite side as $1$ (since $\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = x$ ).

STEP 4

Using the Pythagorean theorem, we can find the hypotenuse $h$ of the right triangle with sides $x$ and $1$ .
$h^2 = x^2 + 1^2$

STEP 5

Solve for the hypotenuse $h$ .
$h = \sqrt{x^2 + 1}$

STEP 6

Now, we can express the secant of $\theta$ in terms of $x$ and $h$ . Secant is the reciprocal of cosine, and cosine is the adjacent side over the hypotenuse. Therefore:
$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$

STEP 7

Substitute the value of $h$ into the expression for $\sec \theta$ .
$\sec \theta = \frac{\sqrt{x^2 + 1}}{x}$

STEP 8

Square the expression to find $\sec^2 \theta$ .
$\sec^2 \theta = \left(\frac{\sqrt{x^2 + 1}}{x}\right)^2$

STEP 9

Simplify the squared expression.
$\sec^2 \theta = \frac{x^2 + 1}{x^2}$

STEP 10

Now, we replace $\theta$ with $\cot^{-1} x$ to find $\sec^2(\cot^{-1} x)$ .
$\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}$

STEP 11

We can see that the expression matches one of the given options.
The solution is:
$\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}$
Therefore, the correct answer is $\frac{1+x^2}{x^2}$ .

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Find the secant squared of the inverse cotangent of xxx. Select the correct answer.

STEP 1

STEP 2

First, let's denote the angle whose cotangent is xxx as θ\thetaθ. Therefore, we have:cot⁡θ=x\cot \theta = xcotθ=x

STEP 3

We know that the cotangent of an angle in a right triangle is the adjacent side over the opposite side. Let's represent the adjacent side as xxx and the opposite side as 111 (since cot⁡θ=adjacentopposite=x\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = xcotθ=oppositeadjacent​=x).

STEP 4

Using the Pythagorean theorem, we can find the hypotenuse hhh of the right triangle with sides xxx and 111.h2=x2+12h^2 = x^2 + 1^2h2=x2+12

STEP 5

Solve for the hypotenuse hhh.h=x2+1h = \sqrt{x^2 + 1}h=x2+1​

STEP 6

Now, we can express the secant of θ\thetaθ in terms of xxx and hhh. Secant is the reciprocal of cosine, and cosine is the adjacent side over the hypotenuse. Therefore:sec⁡θ=1cos⁡θ=hx\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}secθ=cosθ1​=xh​

STEP 7

Substitute the value of hhh into the expression for sec⁡θ\sec \thetasecθ.sec⁡θ=x2+1x\sec \theta = \frac{\sqrt{x^2 + 1}}{x}secθ=xx2+1​​

STEP 8

Square the expression to find sec⁡2θ\sec^2 \thetasec2θ.sec⁡2θ=(x2+1x)2\sec^2 \theta = \left(\frac{\sqrt{x^2 + 1}}{x}\right)^2sec2θ=(xx2+1​​)2

STEP 9

Simplify the squared expression.sec⁡2θ=x2+1x2\sec^2 \theta = \frac{x^2 + 1}{x^2}sec2θ=x2x2+1​

STEP 10

Now, we replace θ\thetaθ with cot⁡−1x\cot^{-1} xcot−1x to find sec⁡2(cot⁡−1x)\sec^2(\cot^{-1} x)sec2(cot−1x).sec⁡2(cot⁡−1x)=x2+1x2\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}sec2(cot−1x)=x2x2+1​

STEP 11

We can see that the expression matches one of the given options.The solution is:sec⁡2(cot⁡−1x)=x2+1x2\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}sec2(cot−1x)=x2x2+1​Therefore, the correct answer is 1+x2x2\frac{1+x^2}{x^2}x21+x2​.

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Find the secant squared of the inverse cotangent of $x$ . Select the correct answer.

First, let's denote the angle whose cotangent is $x$ as $\theta$ . Therefore, we have:
$\cot \theta = x$

We know that the cotangent of an angle in a right triangle is the adjacent side over the opposite side. Let's represent the adjacent side as $x$ and the opposite side as $1$ (since $\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = x$ ).

Using the Pythagorean theorem, we can find the hypotenuse $h$ of the right triangle with sides $x$ and $1$ .
$h^2 = x^2 + 1^2$

Solve for the hypotenuse $h$ .
$h = \sqrt{x^2 + 1}$

Now, we can express the secant of $\theta$ in terms of $x$ and $h$ . Secant is the reciprocal of cosine, and cosine is the adjacent side over the hypotenuse. Therefore:
$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$

Substitute the value of $h$ into the expression for $\sec \theta$ .
$\sec \theta = \frac{\sqrt{x^2 + 1}}{x}$

Square the expression to find $\sec^2 \theta$ .
$\sec^2 \theta = \left(\frac{\sqrt{x^2 + 1}}{x}\right)^2$

Simplify the squared expression.
$\sec^2 \theta = \frac{x^2 + 1}{x^2}$

Now, we replace $\theta$ with $\cot^{-1} x$ to find $\sec^2(\cot^{-1} x)$ .
$\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}$

We can see that the expression matches one of the given options.
The solution is:
$\sec^2(\cot^{-1} x) = \frac{x^2 + 1}{x^2}$
Therefore, the correct answer is $\frac{1+x^2}{x^2}$ .