# Question 75094

Apr 16, 2017

$\csc \left(2 \arctan \left(\frac{3}{4}\right)\right) = \frac{25}{24}$

#### Explanation:

Because $\sin$ and $\csc$ are reciprocals:

$\csc \left(2 \arctan \left(\frac{3}{4}\right)\right) = \frac{1}{\sin} \left(2 \arctan \left(\frac{3}{4}\right)\right)$

Using the double angle identity $\sin \left(2 \theta\right) = 2 \sin \left(\theta\right) \cos \left(\theta\right)$:

=1/(2color(blue)(sin(arctan(3/4)))color(red)(cos(arctan(3/4)))#

We can find the values of $\sin \left(\arctan \left(\frac{3}{4}\right)\right)$ and $\cos \left(\arctan \left(\frac{3}{4}\right)\right)$ using a similar method.

Note that when $\theta = \arctan \left(\frac{3}{4}\right)$, then $\tan \left(\theta\right) = \frac{3}{4}$. That is, where $\theta$ is an angle in a right triangle, the side opposite $\theta$ is $3$ and the leg adjacent to $\theta$ is $4$. The Pythagorean theorem tells us that the hypotenuse is $5$.

We then see that:

$\textcolor{b l u e}{\sin \left(\arctan \left(\frac{3}{4}\right)\right)} = \sin \left(\theta\right) = \text{opposite"/"hypotenuse} = \frac{3}{5}$

$\textcolor{red}{\cos \left(\arctan \left(\frac{3}{4}\right)\right)} = \cos \left(\theta\right) = \text{adjacent"/"hypotenuse} = \frac{4}{5}$

So the original expression is:

$= \frac{1}{2 \textcolor{b l u e}{\left(\frac{3}{5}\right)} \textcolor{red}{\left(\frac{4}{5}\right)}}$

$= \frac{25}{24}$