Introduction

The reciprocals of sine, cosine, and tangent are given special names:

  • The reciprocal of sine is cosecant, abbreviated \csc. \csc(\theta) = \dfrac 1 {\sin(\theta)} = \dfrac 1 {\dfrac {\text{opposite}} {\text{hypotenuse}}} = \dfrac {\text{hypotenuse}} {\text{opposite}}

  • The reciprocal of cosine is secant, abbreviated \sec. \sec (\theta) = \dfrac 1 {\cos(\theta)} = \dfrac 1 {\dfrac {\text{adjacent}} {\text{hypotenuse}}} = \dfrac {\text{hypotenuse}} {\text{adjacent}}

  • The reciprocal of tangent is cotangent, abbreviated \cot. \cot (\theta) = \dfrac 1 {\tan(\theta)} = \dfrac 1 {\dfrac {\text{opposite}} {\text{adjacent}}} = \dfrac {\text{adjacent}} {\text{opposite}}

Let's write down expressions for \csc(B), \sec(B), and \cot(B) for the triangle below.



\eqalign{ \csc(B) &= \dfrac {\text{hypotenuse}} {\text{opposite}} = \dfrac c b \\[5pt] \sec(B) &= \dfrac {\text{hypotenuse}} {\text{adjacent}} = \dfrac c a \\[5pt] \cot(B) &= \dfrac {\text{adjacent}} {\text{opposite}} = \dfrac a b \\ }

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Example: Evaluating Reciprocal Trigonometric Ratios Given a Triangle

Find \csc\theta, \sec\theta, and \cot\theta for the triangle below.


EXPLANATION

We compute the three reciprocal trigonometric ratios as follows:

\eqalign{ \csc\theta &= \dfrac{1}{\sin\theta} = \dfrac {\text{hypotenuse}} {\text{opposite}} = \dfrac 5 3, \\[5pt] \sec\theta &= \dfrac{1}{\cos\theta} = \dfrac {\text{hypotenuse}} {\text{adjacent}} = \dfrac 5 4, \\[5pt] \cot\theta &= \dfrac{1}{\tan\theta} = \dfrac {\text{adjacent}} {\text{opposite}} = \dfrac 4 3. }

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Practice: Evaluating Reciprocal Trigonometric Ratios Given a Triangle

$\cot{Q} =$

a
 $ \dfrac{24}{7}$
b
 $\dfrac{24}{25}$
c
 $\dfrac{25}{7}$
d
 $\dfrac{7}{25}$
e
 $\dfrac{7}{24}$
Practice: Evaluating Reciprocal Trigonometric Ratios Given a Triangle

$\sec{S} =$

a
 $\dfrac{4}{3}$
b
 $\dfrac{3}{5}$
c
 $\dfrac{3}{4}$
d
 $ \dfrac{5}{3}$
e
 $\dfrac{4}{5}$
The Reciprocal Trigonometric Ratios and the Pythagorean Theorem

If we are given two sides of a right triangle, how can we calculate the reciprocal ratios \sec, \csc, or \cot when we are missing one of the necessary sides in the ratio?

The trick here is the same that we use for finding \sin, \cos, or \tan\mathbin{:} we use the Pythagorean theorem for finding the missing side.

To illustrate, let's calculate \csc L in the triangle below.

Remember that \csc is the reciprocal of \sin. So first we need to compute sine:

\sin L = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{JK}{KL} = \dfrac{JK}{4}

To find the length JK, we can use the Pythagorean Theorem:

\begin{align} JK &= \sqrt{ KL^2 - JL^2 } \\ &= \sqrt{ 4^2 - 3^2 } \\ &= \sqrt{16-9 } \\ &= \sqrt{7} \end{align}

So we have

\sin L = \dfrac{JK}{4} = \dfrac{ \sqrt{7} }{4},

and therefore

\begin{align} \csc L &= \left( \dfrac{ \sqrt{7} }{4} \right)^{-1} \\[2pt] &= \dfrac{4}{\sqrt{7}} \\[2pt] &= \dfrac{4 \cdot \sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} \\[2pt] &= \dfrac{4\sqrt{7}}{7} . \end{align}

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Example: Evaluating a Reciprocal Trigonometric Ratio Given a Triangle Using the Pythagorean Theorem

Find \sec L for \triangle KLJ (shown below).

EXPLANATION

Remember that \sec is the reciprocal of \cos. So first we need to compute cosine:

\cos L = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{JL}{KL} = \dfrac{JL}{5}

To find the length JL, we can use the Pythagorean Theorem:

\begin{align} JL &= \sqrt{ KL^2 - JK^2 } \\ &= \sqrt{ 5^2 - 3^2 } \\ &= \sqrt{25-9 } \\ &= \sqrt{16} \\ &= 4 \end{align}

So we have

\cos L = \dfrac{JL}{5} = \dfrac{4}{5},

and therefore

\begin{align} \sec L &= \left( \dfrac{ 4 }{5} \right)^{-1} = \dfrac{5}{4}. \end{align}

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Practice: Evaluating a Reciprocal Trigonometric Ratio Given a Triangle Using the Pythagorean Theorem

$\csc A =$

a
 $\dfrac{8}{15}$
b
 $\dfrac{17}{8}$
c
 $\dfrac{17}{15}$
d
 $\dfrac{15}{17}$
e
 $\dfrac{15}{8}$
Practice: Evaluating a Reciprocal Trigonometric Ratio Given a Triangle Using the Pythagorean Theorem

$\cot{Z} =$

a
 $\dfrac{4}{3}$
b
 $\dfrac{5}{4}$
c
 $ \dfrac{3}{4}$
d
 $\dfrac{3}{5}$
e
 $\dfrac{4}{5}$
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