Question

If $\sin \theta =\dfrac{5}{13}$, $\theta $ being in quadrant II, how do you find the exact value of each of the remaining trigonometric functions of $\theta $ ?

Answer 1

Hint:We explain the function $\sin \theta =\dfrac{5}{13}$ and the quadrant value for the angle $\theta $. We express the identity functions of other ratios with ratio of sin. It’s given that $\sin \theta =\dfrac{5}{13}$. Thereafter we put the value to find the value of each of the remaining trigonometric functions. We also use the representation of a right-angle triangle with height and hypotenuse ratio being $\dfrac{12}{13}$.

Complete step by step answer:
It’s given that $\sin \theta =\dfrac{5}{13}$, $\theta $ being in quadrant II. In that quadrant only $\sin \theta $ and $\csc \theta $ are positive whereas all the other ratios are negative.
We know the sum of the square law of ${{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1$.
We can put the value of $\sin \theta =\dfrac{5}{13}$ in the equation of ${{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1$.
Putting the value of $\sin \theta $, we get ${{\left( \dfrac{5}{13} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1$.
Now we perform the binary operations.
${{\left( \dfrac{5}{13} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1 \\
\Rightarrow {{\left( \cos \theta \right)}^{2}}=1-\dfrac{25}{169}=\dfrac{144}{169} \\
\Rightarrow \left( \cos \theta \right)=\pm \dfrac{12}{13} \\
$
The value of $\cos \theta $ in quadrant II will be negative and that’s why $\left( \cos \theta \right)=-\dfrac{12}{13}$.
We can find the identity which gives $\tan x=\dfrac{\sin x}{\cos x}$.
Putting the values we get,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }\\
\Rightarrow\tan \theta =\dfrac{\dfrac{5}{13}}{-\dfrac{12}{13}}\\
\Rightarrow\tan \theta =-\dfrac{5}{12}\\$
We can find the value of $\sec \theta ,\csc \theta ,\cot \theta $ from the relation of \[\left( \sec x \right)=\dfrac{1}{\cos x},\left( \csc x \right)=\dfrac{1}{\sin x},\left( \cot x \right)=\dfrac{1}{\tan x}\].
Putting the value, we get,
\[\Rightarrow \left( \sec \theta \right)=\dfrac{1}{\cos \theta }=-\dfrac{13}{12}\]
\[\Rightarrow \left( \csc \theta \right)=\dfrac{1}{\sin \theta }=\dfrac{13}{5}\]
\[\Rightarrow \left( \cot \theta \right)=\dfrac{1}{\tan \theta }=-\dfrac{12}{5}\]

Hence, in this way we have found the values of all other trigonometric functions.

Note: In addition to the reciprocal relationships of certain trigonometric functions, two other types of relationships exist. These relationships, known as trigonometric identities, include functional relationships and Pythagorean relationships.