Because [latex]-\dfrac{\pi}{2} \le \theta \le 0{,}[/latex] we draw a reference triangle in the fourth quadrant, as shown at right. Because [latex]\sec \theta = 3 = \dfrac{3}{1}{,}[/latex] we label the horizontal leg with [latex]x=1[/latex] and the hypotenuse with [latex]r=3{.}[/latex]

From the Pythagorean theorem, we find [latex]y=-\sqrt{8}=-2\sqrt{2}{.}[/latex] We can now compute the values of the six trigonometric ratios.

With the calculator in degree mode, enter
[latex]\qquad\qquad\qquad[/latex] 1 ÷ COS 47 ) ENTER
to obtain [latex]\sec 47° \approx 1.466{.}[/latex] Or we can calculate [latex]\cos 47°[/latex] first, and then use the reciprocal key:
[latex]\qquad\qquad\qquad[/latex]COS 47 ) ENTER [latex]\small\boxed{x^{-1}}[/latex] ENTER

The cosecant is undefined when its denominator, [latex]\sin \theta{,}[/latex] equals zero, and [latex]\sin \theta = 0[/latex] when [latex]\theta[/latex] is a multiple of [latex]180°{.}[/latex] In radians, [latex]\csc \theta[/latex] is undefined if [latex]\theta[/latex] is a multiple of [latex]\pi{.}[/latex]

1. From the figure, we see that [latex]\dfrac{L}{h}=\cot \theta{,}[/latex] or [latex]L=h\cot \theta{.}[/latex]
2. Substituting [latex]{3}[/latex] for [latex]h[/latex] and [latex]{20°}[/latex] for [latex]\theta{,}[/latex] we find
   [latex]L = {3} \cot {20°} = 3(2.7475)=8.24[/latex]
   The shadow is about 8.24 meters long.

Consider the graph of [latex]y=\cos x[/latex] shown at left below. When [latex]x=\dfrac{-\pi}{2},~\dfrac{\pi}{2}[/latex] and [latex]\dfrac{3\pi}{2},~ \cos x=0{,}[/latex] so [latex]\sec x[/latex] is undefined at these [latex]x[/latex]-values, and we insert vertical asymptotes at those [latex]x[/latex]-values to start our graph of [latex]y=\sec x{,}[/latex] as shown at right below.


To find some points on the graph, we look at points on the graph of [latex]y=\cos x{.}[/latex] At each [latex]x[/latex]-value, the [latex]y[/latex]-coordinate of the point on the graph of [latex]y=\sec x[/latex] is the reciprocal of [latex]\cos x{.}[/latex]
For example, at [latex]x=0[/latex] and [latex]x=2\pi{,}[/latex] we have [latex]\cos x = 1{,}[/latex] so [latex]\sec x = \frac{1}{1} = 1{.}[/latex] Thus, we plot the points [latex](0,1)[/latex] and [latex](2\pi,1)[/latex] on the graph of [latex]f(x)=\sec x{.}[/latex] Similarly, at [latex]x=-\pi[/latex] and [latex]x=\pi{,}[/latex] [latex]\cos x = -1{,}[/latex] so the value of [latex]\sec x[/latex] is [latex]\frac{1}{-1} = -1{,}[/latex] and we plot the points [latex](-\pi,-1)[/latex] and [latex](\pi,-1)[/latex] on the graph of [latex]f(x)=\sec x{.}[/latex]
Finally, we notice that the values of [latex]\cos x[/latex] are decreasing toward [latex]0[/latex] as [latex]x[/latex] increases from [latex]0[/latex] to [latex]\dfrac{\pi}{2}{,}[/latex] so the graph of [latex]f(x)=\sec x[/latex] increases toward [latex]\infty[/latex] on the same interval. By similar arguments, we fill in the graph of [latex]f(x)=\sec x[/latex] between each of the vertical asymptotes to produce the graph below.

Because the cosine is defined for all real numbers, the domain of the secant includes all real numbers except for values where the cosine is zero. These values are the odd multiples of [latex]\dfrac{\pi}{2}{;}[/latex] that is, [latex]\dfrac{\pi}{2},~\dfrac{3\pi}{2},~\dfrac{5\pi}{2},~ \ldots{,}[/latex] and their opposites. Because the range of the cosine consists of all [latex]y[/latex]-values with [latex]-1 \le y \le 1{,}[/latex] the range of the secant includes the reciprocals of those values; namely, [latex]y \ge 1[/latex] and [latex]y \le -1{.}[/latex]

We take the reciprocal of each side of the equation to obtain

[latex]\sin \theta = \dfrac{3}{2\sqrt{3}} = \dfrac{\sqrt{3}}{2}[/latex]

Because [latex]\dfrac{\sqrt{3}}{2}[/latex] is one of the special values, we recognize that one of the solutions is [latex]\theta =\dfrac{\pi}{3}{.}[/latex] The sine and the cosecant are also positive in the second quadrant, so the second solution is [latex]\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}{.}[/latex]

Because [latex]\sec \theta = \dfrac{1}{\cos \theta}{,}[/latex] we see that [latex]\dfrac{1}{\cos \theta}=3{,}[/latex] or [latex]\cos \theta = \dfrac{1}{3}{.}[/latex] We

Because [latex]\theta[/latex] lies in the fourth quadrant, where the sine function is negative, we choose the negative square root for [latex]\sin \theta{.}[/latex] Once we know [latex]\sin \theta{,}[/latex] we calculate its reciprocal to find [latex]\csc\theta{.}[/latex]

[latex]\sin \theta = -\sqrt{\dfrac{8}{9}} = \dfrac{-2\sqrt{2}}{3},~{ and }~\csc\theta = \dfrac{1}{\sin \theta} = \dfrac{-3}{2\sqrt{2}} = \dfrac{-3\sqrt{2}}{4}[/latex]

Because the sine is the reciprocal of the cosecant, we have [latex]\sin x = \dfrac{1}{\csc x} = \dfrac{1}{w}{.}[/latex] We substitute [latex]{\dfrac{1}{w}}[/latex] for [latex]\sin x[/latex] in the Pythagorean identity to find

[latex]\cos x = \pm \sqrt{1 - \sin^2 x} = \pm \sqrt{1 - \left({\dfrac{1}{w}}\right)^2}[/latex]

We choose the positive root because cosine is positive in the first quadrant and simplify to get

[latex]\cos x = \sqrt{1 - \dfrac{1}{w^2}} = \sqrt{\dfrac{w^2-1}{w^2}} = \dfrac{\sqrt{w^2 - 1}}{|w|}[/latex]

We can replace [latex]|w|[/latex] by [latex]w[/latex] in this last expression because [latex]w \gt 0{.}[/latex] (Do you see why [latex]w \gt 0{?}[/latex]) Finally, because the cotangent is the reciprocal of the tangent, we have

[latex]\cot x = \dfrac{\cos x}{\sin x} = \dfrac{\dfrac{w^2-1}{w}}{\dfrac{1}{w}}=\sqrt{w^2 - 1}[/latex]

We replace [latex]~\sec \theta~[/latex] by [latex]~\dfrac{1}{\cos \theta}~[/latex] and [latex]~\tan \theta~[/latex] by [latex]~\dfrac{\sin \theta}{\cos \theta}~[/latex] to get

[latex]\dfrac{1}{\cos \theta} - \dfrac{\sin \theta}{\cos \theta} \cdot \sin \theta = \dfrac{1}{\cos \theta} - \dfrac{\sin^2 \theta}{\cos \theta}\\ = \dfrac{1 - \sin^2 \theta}{\cos \theta}= \dfrac{\cos^2 \theta}{\cos \theta} = \cos \theta[/latex]

We cannot find the sine and cosine of an angle directly from the value of the tangent; in particular, it is not true that [latex]\sin \alpha = 3[/latex] and [latex]\cos \alpha = 5{!}[/latex] (Do you see why?) Instead, we begin with the Pythagorean identity for the tangent.

[latex]\sec^2 \alpha = 1 + \tan^2 \alpha = 1 + (\dfrac{3}{5})^2\\ = \dfrac{25}{25} + \dfrac{9}{25} = \dfrac{34}{25}\\ \sec \alpha = \pm \sqrt{\dfrac{34}{25}} = \dfrac{\pm \sqrt{34}}{5}[/latex]

Because [latex]\alpha[/latex] is in the third quadrant, both its sine and cosine are negative. Therefore the reciprocals of cosine and sine—namely, secant and cosecant—must also be negative, and hence [latex]\sec \alpha = \dfrac{-\sqrt{34}}{5}{.}[/latex] The cosine of [latex]\alpha[/latex] is the reciprocal of the secant, so [latex]\cos \alpha = \dfrac{-5}{\sqrt{34}}{.}[/latex]