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

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

Question
Trigonometric Functions
asked 2021-01-04


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

Answers (1)

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.
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