Question

If $3\tan \theta =1$, find $\sin \theta$, $\cos \theta$ and $\cot \theta$.

Answer 1

Let $△ABC$ be a right angled triangle where $\angle B={90}^0$ and $\angle C=\theta$ as shown in the above figure:

Now it is given that $3\tan \theta =1$ or $\tan \theta =\frac{1}{3}$ and we know that, in a right angled triangle, $\tan \theta$ is equal to opposite side over adjacent side that is $\tan \theta =\frac{Oppositeside}{Adjacentside}$, therefore, opposite side $AB=1$ and adjacent side $BC=3$.

Now, using pythagoras theorem in $△ABC$, we have

$A{C}^2=A{B}^2+B{C}^2=1^2+3^2=1+9=10\Rightarrow AC=\sqrt{10}$

Therefore, the hypotenuse $AC=\sqrt{10}$.

We know that, in a right angled triangle,

$\sin \theta$ is equal to opposite side over hypotenuse that is $\sin \theta =\frac{Oppositeside}{Hypotenuse}$and,

$\cos \theta$ is equal to adjacent side over hypotenuse that is $\cos \theta =\frac{Adjacentside}{Hypotenuse}$

Here, we have opposite side $AB=\sqrt{3}$, adjacent side $BC=1$ and the hypotenuse $AC=2$, therefore, $\cos \theta$, $\sin \theta$ and$\cot \theta$ can be determined as follows:

$\sin \theta =\frac{Oppositeside}{Hypotenuse}=\frac{AB}{AC}=\frac{1}{\sqrt{10}}$

$\cos \theta =\frac{Adjacentside}{Hypotenuse}=\frac{BC}{AC}=\frac{3}{\sqrt{10}}$

$\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\frac{1}{3}}=1\times \frac{3}{1}=3$

Hence, $\sin \theta =\frac{1}{\sqrt{10}}$ and $\cos \theta =\frac{3}{\sqrt{10}}$ and $\cot \theta =3$.