Reciprocal identities

In addition to the sine, cosine and tangent functions, three further trigonometric functions are commonly used. They are the functions cosecant, secant and cotangent, defined as follows:

$\csc\theta\equiv\frac{1}{\sin\theta}$cscθ≡1sinθ​

$\sec\theta\equiv\frac{1}{\cos\theta}$secθ≡1cosθ​

$\cot\theta\equiv\frac{1}{\tan\theta}$cotθ≡1tanθ​

The functions sine and cosecant are said to be reciprocals of one another, as are the pairs cosine and secant, and tangent and cotangent. It follows that, for example, if $\sin\theta=\frac{a}{b}$sinθ=ab​, then $\csc\theta=\frac{b}{a}$cscθ=ba​. 

If we compare the behaviours of reciprocal pairs of trigonometric functions we notice that when one of the function values approaches zero, the value of its reciprocal function becomes arbitrarily large. Moreover, when one function attains the value zero, the reciprocal ceases to exist - because the fraction $\frac{1}{0}$10​ is undefined. This occurs, for example, when $\cos\theta=\frac{\pi}{2}$cosθ=π2​ or when $\cos\theta=\frac{3\pi}{2}$cosθ=3π2​.

Because the functions sine, cosine and tangent are periodic, the reciprocal functions cosecant, secant and cotangent are also periodic, with the same periods as their respective originals.

Using the identities that we now possess, it is possible to derive the values of all six of the trigonometric functions if we are given the value of one of them.

Example

Given that $\sin\beta=\frac{5}{9}$sinβ=59​, deduce the values of the other five trigonometric functions at the same value of $\beta$β.

We can use the identity $\sin^2\beta+\cos^2\beta\equiv1$sin2β+cos2β≡1 to deduce $\cos\beta$cosβ.  Hence, $\left(\frac{5}{9}\right)^2+\cos^2\beta=1$(59​)2+cos2β=1 and so,

$\cos\beta=\sqrt{1-\left(\frac{5}{9}\right)^2}=\frac{2\sqrt{14}}{9}$cosβ=√1−(59​)2=2√149​

Now,

$\csc\beta=\frac{1}{\sin\beta}=\frac{9}{5},$cscβ=1sinβ​=95​,

$\sec\beta=\frac{1}{\cos\beta}=\frac{9}{2\sqrt{14}}=\frac{9\sqrt{14}}{28},$secβ=1cosβ​=92√14​=9√1428​,

$\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{5}{2\sqrt{14}}=\frac{5\sqrt{14}}{28},$tanβ=sinβcosβ​=52√14​=5√1428​,

$\cot\beta=\frac{1}{\tan\beta}=\frac{2\sqrt{14}}{5}$cotβ=1tanβ​=2√145​

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3