Question

How do you find the values of the six trigonometric functions given $\mathrm{cosec\theta }=4$ and $\mathrm{cot\theta }<0$?

Answer 1

Step- 1: Formula used:

Trigonometric identity

${\mathrm{cosec}}^2\mathrm{\theta }=1+{\cot }^2\mathrm{\theta }$

$\mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}$

Step-2: Calculate the value of trigonometric functions

Given that,

$$\begin{gathered} \mathrm{cosec\theta }=4 \\ \mathrm{cot\theta }<0 \end{gathered}$$

Thus,

${\mathrm{cosec}}^2\mathrm{\theta }=1+{\cot }^2\mathrm{\theta }$

$\mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}$

$4^2=1+{\cot }^2\theta$

${\cot }^2\theta =15$ and as $\mathrm{cot\theta }<0,\mathrm{cot\theta }=-\sqrt{15}$

So $\mathrm{tan\theta }=-\frac{1}{\sqrt{15}}\mathrm{and}\mathrm{sin\theta }=\frac{1}{4}$

As $\mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}$

$$\begin{gathered} \mathrm{cos\theta }=\frac{\mathrm{sin\theta }}{\mathrm{tan\theta }} \\ \mathrm{cos\theta }=\frac{{\displaystyle \frac{1}{4}}}{-{\displaystyle \frac{1}{\sqrt{15}}}} \end{gathered}$$

$\mathrm{cos\theta }=-\frac{\sqrt{15}}{4}$

and $\mathrm{sec\theta }=-\frac{4}{\sqrt{15}}$

Therefore the values of trigonometric functions are

$\mathrm{sin\theta }=\frac{1}{4},\mathrm{cos\theta }=-\frac{\sqrt{15}}{4},\mathrm{tan\theta }=-\frac{1}{\sqrt{15}},\mathrm{cot\theta }=-\sqrt{15},\mathrm{sec\theta }=-\frac{4}{\sqrt{15}},\mathrm{cosec\theta }=4$