# How do you evaluate tan(csc^-1(sqrt7/2)) without a calculator?

Aug 13, 2016

$\tan \left({\csc}^{- 1} \left(\frac{\sqrt{7}}{2}\right)\right) = \frac{2 \sqrt{3}}{3}$

#### Explanation:

Consider a triangle with sides $2$, $\sqrt{3}$ and $\sqrt{7}$.

It is a right angled triangle, since:

${2}^{2} + {\left(\sqrt{3}\right)}^{2} = 4 + 3 = 7 = {\left(\sqrt{7}\right)}^{2}$

Since $\csc \theta = \frac{1}{\sin} \theta = \text{hypotenuse"/"opposite}$, and $\tan \theta = \text{opposite"/"adjacent}$ we have:

$\text{hypotenuse} = \sqrt{7}$

$\text{opposite} = 2$

$\text{adjacent} = \sqrt{3}$

$\tan \left({\csc}^{- 1} \left(\frac{\sqrt{7}}{2}\right)\right) = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

Aug 13, 2016

$= \frac{2 \sqrt{3}}{3}$

#### Explanation:

Let $\theta = {\csc}^{-} 1 \left(\frac{\sqrt{7}}{2}\right)$
or
$\csc \theta = \frac{\sqrt{7}}{2}$
or
$\frac{1}{\sin} \theta = \frac{\sqrt{7}}{2}$
or
$\sin \theta = \frac{2}{\sqrt{7}}$
or
$\theta = {\sin}^{-} 1 \left(\frac{2}{\sqrt{7}}\right)$
Hence
$\frac{p}{h} = \frac{2}{\sqrt{7}}$
$\cos \theta = \frac{b}{h}$
or
$\cos \theta = \frac{\sqrt{{h}^{2} - {p}^{2}}}{h}$
or
$\cos \theta = \frac{\sqrt{{\left(\sqrt{7}\right)}^{2} - {2}^{2}}}{\sqrt{7}}$
or
$\cos \theta = \frac{\sqrt{7 - 4}}{\sqrt{7}}$
or
$\cos \theta = \frac{\sqrt{3}}{\sqrt{7}}$
Therefore
$\tan \theta = \sin \frac{\theta}{\cos} \theta$
or
$\tan \theta = \frac{\frac{2}{\sqrt{7}}}{\frac{\sqrt{3}}{\sqrt{7}}}$
or
$\tan \theta = \frac{2}{\sqrt{3}}$
or
$\tan \left({\csc}^{-} 1 \left(\frac{\sqrt{7}}{2}\right)\right) = \frac{2}{\sqrt{3}}$
or
$\tan \left({\csc}^{-} 1 \left(\frac{\sqrt{7}}{2}\right)\right) = \frac{2 \sqrt{3}}{3}$

Sep 25, 2016

$\tan \left({\csc}^{-} 1 \left(\frac{\sqrt{7}}{2}\right)\right)$

$= \tan \left({\cot}^{-} 1 \left(\sqrt{{\left(\frac{\sqrt{7}}{2}\right)}^{2} - 1}\right)\right)$

$= \tan \left({\cot}^{-} 1 \sqrt{\frac{7}{4} - 1}\right)$

$= \tan \left({\cot}^{-} 1 \sqrt{\frac{3}{4}}\right)$

$= \tan \left({\tan}^{-} 1 \left(\frac{2}{\sqrt{3}}\right)\right) = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$