Answer
The required value of $\tan \frac{7\pi }{8}$ is $-\sqrt{2}+1$.
Work Step by Step
We have to find the value of $\tan \frac{7\pi {}^\circ }{8}$; the formula is used which represents the tan identity in term of cos and sine. The term $\tan \frac{7\pi {}^\circ }{8}$ comes under the second quadrant where the value of sine and cos function is positive and the rest of the trigonometric functions are negative.
$\tan \frac{7\pi {}^\circ }{8}=\tan \frac{\frac{7\pi }{4}}{2}=\frac{1-\cos \frac{7\pi }{4}}{\sin \frac{7\pi }{4}}=\frac{1-\left( \frac{\sqrt{2}}{2} \right)}{-\frac{\sqrt{2}}{2}}=\frac{1}{-\frac{\sqrt{2}}{2}}-\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}$
So in order to solve the problem, write the denominator with each respective number.
$\begin{align}
& \tan \frac{7\pi {}^\circ }{8}=\frac{-2}{\sqrt{2}}+1 \\
& =\frac{-2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}+1 \\
& =-\sqrt{2}+1
\end{align}$
Then, the rationalizing technique is used to compute the value in the aforementioned situation.
Thus, the value of $\tan \frac{7\pi }{8}$ is $-\sqrt{2}+1$.
