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.