Question

# sec theta = -3, tan theta > 0. Find the exact value of the remaining trigonometric functions of theta.

Trigonometric Functions

$$\sec \theta = -3, \tan \theta > 0$$. Find the exact value of the remaining trigonometric functions of
$$\theta$$.

2021-01-05
We have given the value of a trigonometric function $$\displaystyle{\sec{{\left(\theta\right)}}}$$ and asked the values of other remaining trigonometric functions. So here we will use some standard identities and results in order to determine the values of other trigonometric functions.
$$\displaystyle{{\sec}^{{2}}\theta}-{{\tan}^{{2}}\theta}={1}$$
$$\displaystyle{\cos{\theta}}=\frac{{1}}{{{\sec{\theta}}}}$$
$$\displaystyle{\sin{\theta}}=\frac{{{\tan{\theta}}}}{{{\sec{\theta}}}}$$
$$\displaystyle{\cos{{e}}}{c}\theta=\frac{{1}}{{{\sin{\theta}}}}$$
$$\displaystyle{\cot{\theta}}=\frac{{1}}{{{\tan{\theta}}}}$$
Since, $$\displaystyle{\sec{\theta}}=−{3}{<}{0}{\quad\text{and}\quad}{\tan{\theta}}{>}{0}$$ then $$\displaystyle\theta$$ must be in third quadrant in which $$\displaystyle{\sin{\theta}},{\cos{\theta}},{\cos{{e}}}{c}\theta{\quad\text{and}\quad}{\sec{\theta}}$$ will be negative and rest $$\displaystyle{\tan{\theta}}{\quad\text{and}\quad}{\cot{\theta}}$$ will be positive.
Now,
$$\displaystyle{{\sec}^{{2}}\theta}-{{\tan}^{{2}}\theta}={1}$$
$$\displaystyle{\left(-{3}\right)}^{{2}}-{{\tan}^{{2}}\theta}={1}$$
$$\displaystyle{{\tan}^{{2}}\theta}={\left(-{3}\right)}^{{2}}-{1}$$
$$\displaystyle{\tan{\theta}}=\sqrt{{8}}$$
Also,
$$\displaystyle{\cos{\theta}}=\frac{{1}}{{{\sec{\theta}}}}=\frac{{1}}{{-{{3}}}}=-\frac{{1}}{{3}}$$
$$\displaystyle{\sin{\theta}}=\frac{{{\tan{\theta}}}}{{{\sec{\theta}}}}=\frac{\sqrt{{8}}}{{-{{3}}}}=\frac{{-{2}\sqrt{{2}}}}{{3}}$$
$$\displaystyle{\cos{{e}}}{c}\theta=\frac{{1}}{{{\sin{\theta}}}}=\frac{{1}}{{\frac{{-{2}\sqrt{{2}}}}{{3}}}}=-\frac{{3}}{{{2}\sqrt{{2}}}}$$
$$\displaystyle{\cot{\theta}}=\frac{{1}}{{{\tan{\theta}}}}=\frac{{1}}{\sqrt{{8}}}=\frac{{1}}{{{2}\sqrt{{2}}}}$$
Hence, values of other trigonometric functions $$\displaystyle{\sin{\theta}},{\cos{\theta}},{\tan{\theta}}{\cos{{e}}}{c}\theta{\quad\text{and}\quad}{\cot{\theta}}$$ will be $$\displaystyle\frac{{-{2}\sqrt{{2}}}}{{3}},-\frac{{1}}{{3}},{2}\sqrt{{2}},-\frac{{3}}{{{2}\sqrt{{2}}}}{\quad\text{and}\quad}\frac{{1}}{{{2}\sqrt{{2}}}}$$ respectively.