Sec, Csc, Ctan - Trigonometry

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Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

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The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = $\frac{hypotenuse}{adjacent}$ = $\frac{25}{24}$$

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

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Since $\csc\theta=\frac{1}{\sin\theta}$$ :

$$\frac{1}{\csc\theta}$=\frac{1}{\frac{1}${\sin\theta}}=1*$\frac{\sin\theta}{1}$=\sin\theta$

For the above triangle, what is $\sec (\theta)$ if , and ?

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Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = $\frac{1}{\cos\left( \theta\right)}$$

It's formula is:

$\sec(\theta) = $\frac{\textup{hypotenuse}$}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = $\frac{17}{15}$ = 1.13$

For the above triangle, what is $\cot (\theta)$ if , and ?

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Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = $\frac{1}{\tan\left( \theta\right)}$$

It's formula is:

$\cot(\theta) = $\frac{\textup{adjacent}$}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = $\frac{24}{18}$ = 1.33$

Evaluate:

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Evaluate each term separately.

$sec(60)= $\frac{1}{cos(60)}$=\frac{1}{0.5}$= 2$

$csc(45)= $\frac{1}{sin(45)}$= $\frac{1}{\frac{\sqrt2}${2}}= $\frac{2}{\sqrt2}$\times$\frac{\sqrt2}{\sqrt2}$= $\frac{2\sqrt2}{2}$=\sqrt2$

$sec(60)-csc(45)=2-$\frac{2\sqrt2}{2}$=2-\sqrt2$

Determine the value of $cot($\frac{\pi}{4}$)$ .

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Rewrite $cot($\frac{\pi}{4}$)$ in terms of sine and cosine.

$cot($\frac{\pi}{4}$)= $\frac{cos(\frac{\pi}{4}$)}{sin($\frac{\pi}{4}$)}=\frac{\frac{\sqrt2}{2}$}{$\frac{\sqrt2}{2}$}=1$

Pick the ratio of side lengths that would give sec C.

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$\sec\theta=\frac{1}{\cos\theta }$$

Find the ratio of Cosine and take the reciprocal.

$\cos\theta=\frac{AC}{CB}$\rightarrow \sec\theta=\frac{1}{\frac{AC}${CB}}=\mathbf{$\frac{CB}{AC}$}$

If $\sin x=\frac{28}{53}$$ , $\cot x=$

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The sine of an angle in a right triangle (that is not the right angle) can be found by dividing the length of the side opposite the angle by the length of the hypotenuse of the triangle.

From this, the length of the side opposite the angle is proportional to 28, and the length of the hypotenuse is proportional to 53.

Without loss of generality, we'll assume that the sides are actually of length 28 and 53, respectively.

We'll use the Pythagorean theorem to determine the length of the adjacent side, which we'll refer to as .

$\begin{align} $y&=$\sqrt{53^2$$-28^2$}$\ &=$\sqrt{2809-784}$\ &=$\sqrt{2025}$\ &=45 \end{align}$

The cotangent of an angle in a right triangle (that is not the right angle) is can be found by dividing the length of the adjacent side by the length of the opposite side.

$\cot x=\frac{45}{28}$$

Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

Tap to see back →

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = $\frac{hypotenuse}{adjacent}$ = $\frac{25}{24}$$

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

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Since $\csc\theta=\frac{1}{\sin\theta}$$ :

$$\frac{1}{\csc\theta}$=\frac{1}{\frac{1}${\sin\theta}}=1*$\frac{\sin\theta}{1}$=\sin\theta$

For the above triangle, what is $\sec (\theta)$ if , and ?

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Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = $\frac{1}{\cos\left( \theta\right)}$$

It's formula is:

$\sec(\theta) = $\frac{\textup{hypotenuse}$}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = $\frac{17}{15}$ = 1.13$

For the above triangle, what is $\cot (\theta)$ if , and ?

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Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = $\frac{1}{\tan\left( \theta\right)}$$

It's formula is:

$\cot(\theta) = $\frac{\textup{adjacent}$}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = $\frac{24}{18}$ = 1.33$

Evaluate:

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Evaluate each term separately.

$sec(60)= $\frac{1}{cos(60)}$=\frac{1}{0.5}$= 2$

$csc(45)= $\frac{1}{sin(45)}$= $\frac{1}{\frac{\sqrt2}${2}}= $\frac{2}{\sqrt2}$\times$\frac{\sqrt2}{\sqrt2}$= $\frac{2\sqrt2}{2}$=\sqrt2$

$sec(60)-csc(45)=2-$\frac{2\sqrt2}{2}$=2-\sqrt2$

Determine the value of $cot($\frac{\pi}{4}$)$ .

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Rewrite $cot($\frac{\pi}{4}$)$ in terms of sine and cosine.

$cot($\frac{\pi}{4}$)= $\frac{cos(\frac{\pi}{4}$)}{sin($\frac{\pi}{4}$)}=\frac{\frac{\sqrt2}{2}$}{$\frac{\sqrt2}{2}$}=1$

Pick the ratio of side lengths that would give sec C.

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$\sec\theta=\frac{1}{\cos\theta }$$

Find the ratio of Cosine and take the reciprocal.

$\cos\theta=\frac{AC}{CB}$\rightarrow \sec\theta=\frac{1}{\frac{AC}${CB}}=\mathbf{$\frac{CB}{AC}$}$

If $\sin x=\frac{28}{53}$$ , $\cot x=$

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The sine of an angle in a right triangle (that is not the right angle) can be found by dividing the length of the side opposite the angle by the length of the hypotenuse of the triangle.

From this, the length of the side opposite the angle is proportional to 28, and the length of the hypotenuse is proportional to 53.

Without loss of generality, we'll assume that the sides are actually of length 28 and 53, respectively.

We'll use the Pythagorean theorem to determine the length of the adjacent side, which we'll refer to as .

$\begin{align} $y&=$\sqrt{53^2$$-28^2$}$\ &=$\sqrt{2809-784}$\ &=$\sqrt{2025}$\ &=45 \end{align}$

The cotangent of an angle in a right triangle (that is not the right angle) is can be found by dividing the length of the adjacent side by the length of the opposite side.

$\cot x=\frac{45}{28}$$

Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

Tap to see back →

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = $\frac{hypotenuse}{adjacent}$ = $\frac{25}{24}$$

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

Tap to see back →

Since $\csc\theta=\frac{1}{\sin\theta}$$ :

$$\frac{1}{\csc\theta}$=\frac{1}{\frac{1}${\sin\theta}}=1*$\frac{\sin\theta}{1}$=\sin\theta$

For the above triangle, what is $\sec (\theta)$ if , and ?

Tap to see back →

Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = $\frac{1}{\cos\left( \theta\right)}$$

It's formula is:

$\sec(\theta) = $\frac{\textup{hypotenuse}$}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = $\frac{17}{15}$ = 1.13$

For the above triangle, what is $\cot (\theta)$ if , and ?

Tap to see back →

Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = $\frac{1}{\tan\left( \theta\right)}$$

It's formula is:

$\cot(\theta) = $\frac{\textup{adjacent}$}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = $\frac{24}{18}$ = 1.33$