Question

Find the values of the other five trigonometric functions if $\sin \theta =\dfrac{3}{5}$, $\theta$ in quadrant I.

Answer 1

Hint:First of all, try to recollect the signs of the various trigonometric ratios in the first quadrant. Now, find $\operatorname{cosec}\theta$ by using $\dfrac{1}{\sin \theta }$, $\cos \theta$ by $\sqrt{1-{{\sin }^{2}}\theta }$, $\sec \theta$ by $\dfrac{1}{\cos \theta }$, $\tan \theta$ by $\dfrac{\sin \theta }{\cos \theta }$ , and $\cot \theta$ by $\dfrac{1}{\tan \theta }$.

Complete step-by-step answer:
In this question, we have to find the values of the other five trigonometric functions if $\sin \theta =\dfrac{3}{5}$, $\theta$ in quadrant I. Before proceeding with this question, let us see the sign of different trigonometric ratios in different quadrants. We have 6 trigonometric ratios and that are $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\operatorname{cosec}\theta$ and $\sec \theta$.
1. In the first quadrant, that is from 0 to ${{90}^{o}}$ or 0 to $\dfrac{\pi }{2}$, all the trigonometric ratios are positive.
2. In the second quadrant, that is from ${{90}^{o}}$ to ${{180}^{o}}$ or $\dfrac{\pi }{2}$ to $\pi$, only $\sin \theta$ and $\operatorname{cosec}\theta$ are positive.
3. In the third quadrant, that is from ${{180}^{o}}$ to ${{270}^{o}}$ or $\pi$ to $\dfrac{3\pi }{2}$, only $\tan \theta$ and $\cot \theta$ are positive.
4. In the fourth quadrant, that is from ${{270}^{o}}$ to ${{360}^{o}}$ or $\dfrac{3\pi }{2}$ to $2\pi$, only $\cos \theta$ and $\sec \theta$ are positive.
This cycle would repeat after ${{360}^{o}}$.

In this figure, A means all are positive, S means $\sin \theta$ and $\operatorname{cosec}\theta$ are positive, T means $\tan \theta$ and $\cot \theta$ are positive and C means $\cos \theta$ and $\sec \theta$ are positive.
Here, we are given that $\sin \theta =\dfrac{3}{5}$ and $\theta$ is in the first quadrant. So, in this quadrant, all the trigonometric ratios would be positive.
We know that $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$
By substituting the value of $\sin \theta =\dfrac{3}{5}$, we get,
$\operatorname{cosec}\theta =\dfrac{1}{\left( \dfrac{3}{5} \right)}$
$\operatorname{cosec}\theta =\dfrac{5}{3}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
By substituting $\sin \theta =\dfrac{3}{5}$, we get,
${{\left( \dfrac{3}{5} \right)}^{2}}+{{\cos }^{2}}\theta =1$
${{\cos }^{2}}\theta =1-{{\left( \dfrac{3}{5} \right)}^{2}}$
${{\cos }^{2}}\theta =1-\dfrac{9}{25}$
${{\cos }^{2}}\theta =\dfrac{16}{25}$
$\cos \theta =\sqrt{\dfrac{16}{25}}$
$\cos \theta =\pm \dfrac{4}{5}$
Here, we take $\cos \theta =\dfrac{4}{5}$ because in the first quadrant all the trigonometric ratios are positive. We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. So, by substituting $\sin \theta =\dfrac{3}{5}$ and $\cos \theta =\dfrac{4}{5}$, we get,
$\tan \theta =\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}$
$\tan \theta =\dfrac{3}{4}$
We also know that $\sec \theta =\dfrac{1}{\cos \theta }$
By substituting $\cos \theta =\dfrac{4}{5}$, we get,
$\sec \theta =\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4}$
We know that $\cot \theta =\dfrac{1}{\tan \theta }$. By substituting the value of $\tan \theta =\dfrac{3}{4}$, we get,
$\cot \theta =\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$
So, if $\sin \theta =\dfrac{3}{5}$ and $\theta$ is in the first quadrant, the other 5 values we get as,
$\cos \theta =\dfrac{4}{5}$
$\sec \theta =\dfrac{5}{4}$
$\operatorname{cosec}\theta =\dfrac{5}{3}$
$\cot \theta =\dfrac{4}{3}$
$\tan \theta =\dfrac{3}{4}$

Note: In this question, we can also find the value of the magnitude of all the trigonometric ratios by constructing a triangle and using $\sin \theta =\dfrac{3}{5}$ in this way.
$\sin \theta =\dfrac{\text{perpendicular}}{\text{hypotenuse}}=\dfrac{3}{5}=\dfrac{AB}{AC}$

From Pythagoras theorem, we get, CB = 4x.
Now, $\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}}=\dfrac{4x}{5x}=\dfrac{4}{5}$
$\tan \theta =\dfrac{AB}{BC}=\dfrac{3x}{4x}=\dfrac{3}{4}$
Similarly, we can find all the other trigonometric ratios.