Question

If $2\sin \theta =\sqrt{3}$, find $\cos \theta$, $\tan \theta$ and $\cot \theta +\csc \theta$.

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 $2\sin \theta =\sqrt{3}$ or $\sin \theta =\frac{\sqrt{3}}{2}$ and we know that, in a right angled triangle, $\sin \theta$ is equal to opposite side over hypotenuse that is $\sin \theta =\frac{\text{Opposite side}}{\text{Hypotenuse}}$, therefore, opposite side $AB=\sqrt{3}$ and hypotenuse $AC=2$.

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

$A{C}^2=A{B}^2+B{C}^2\Rightarrow 2^2=(\sqrt{3}{)}^2+B{C}^2\Rightarrow 4=3+B{C}^2\Rightarrow B{C}^2=4-3=1\Rightarrow BC=\sqrt{1}=1$

Therefore, the adjacent side $BC=1$.

We know that, in a right angled triangle,

$\cos \theta$ is equal to adjacent side over hypotenuse that is $\cos \theta =\frac{\text{Adjacentd side}}{\text{Hypotenuse}}$ and,

$\tan \theta$ is equal to opposite side over adjacent side that is $\tan \theta =\frac{\text{Opposite side}}{\text{Adjacent side}}$

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

$\cos \theta =\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{BC}{AC}=\frac{1}{2}$

$\tan \theta =\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{AB}{BC}=\frac{\sqrt{3}}{1}=\sqrt{3}$

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

$\csc \theta =\frac{1}{\sin \theta }=\frac{1}{\frac{\sqrt{3}}{2}}=1\times \frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

Now, $(\cot \theta +\csc \theta )=\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\frac{\sqrt{3}\times \sqrt{3}}{\sqrt{3}}=\sqrt{3}$

Hence, $\cos \theta =\frac{1}{2}$ and $\tan \theta =\sqrt{3}$ and $(\cot \theta +\csc \theta )=\sqrt{3}$.