Chapter 6, Problem 23RE

To determine

To find: The exact value of each of the remaining trigonometric functions.

Answer to Problem 23RE

The exact values of each of the trigonometric function are,

  $\begin{array}{l}\sin \theta =\frac{1}{\sqrt{5}};\mathrm{cosec}\theta =\sqrt{5};\\ \tan \theta =-\frac{1}{2};\cos \theta =-\frac{2}{\sqrt{5}};\\ \sec \theta =-\frac{\sqrt{5}}{2};\end{array}$

Explanation of Solution

Given information: $\cot \theta =-2,\left[\frac{\pi }{2}<\theta <\pi \right]$

Calculation:

  $\cot \theta =-2,\left[\frac{\pi }{2}<\theta <\pi \right]$

As per the range of $\theta$ the functions are lies in the second quadrant.

  $\begin{array}{l}\therefore \cot \theta =-2\\ \Rightarrow \tan \theta =\frac{1}{\cot \theta }=-\frac{1}{2}\end{array}$

Therefore,

  $\begin{array}{l}\sec \theta =\sqrt{1+{\tan }^2\theta }\\ =\sqrt{1+{\left(-\frac{1}{2}\right)}^2}\\ =\sqrt{1+\frac{1}{4}}\\ =\sqrt{\frac{5}{4}}\\ =\frac{\sqrt{5}}{2}\end{array}$

Hence, $\cos \theta =\frac{1}{\sec \theta }=\frac{1}{\frac{\sqrt{5}}{2}}=\frac{2}{\sqrt{5}}$

Therefore,

  $\begin{array}{l}\mathrm{cosec}\theta =\sqrt{1+{\cot }^2\theta }\\ =\sqrt{1+{\left(-2\right)}^2}\\ =\sqrt{1+4}\\ =\sqrt{5}\end{array}$

Now, $\sin \theta =\frac{1}{\mathrm{cosec}\theta }=\frac{1}{\sqrt{5}}$

Since, in the second quadrant all functions are negative except $\sin \theta$ and $\mathrm{cosec}\theta$ .

Therefore,

  $\begin{array}{l}\sin \theta =\frac{1}{\sqrt{5}};\mathrm{cosec}\theta =\sqrt{5};\\ \tan \theta =-\frac{1}{2};\cos \theta =-\frac{2}{\sqrt{5}};\\ \sec \theta =-\frac{\sqrt{5}}{2};\end{array}$

Hence, the exact values of each of the trigonometric function are,

  $\begin{array}{l}\sin \theta =\frac{1}{\sqrt{5}};\mathrm{cosec}\theta =\sqrt{5};\\ \tan \theta =-\frac{1}{2};\cos \theta =-\frac{2}{\sqrt{5}};\\ \sec \theta =-\frac{\sqrt{5}}{2};\end{array}$