## Precalculus (6th Edition) Blitzer

The required value of $\tan \frac{3\pi }{8}$ is $\sqrt{2}+1$.
We have to find the value of $\tan \frac{3\pi }{8}$; the formula is used which represents the tan identity in term of cos and sine. And the term $\tan \frac{3\pi }{8}$ comes under the first quadrant where the value of all the trigonometric functions is positive. $\tan \frac{3\pi }{8}=\tan \frac{\frac{3\pi }{4}}{2}=\frac{1-\cos \frac{3\pi }{4}}{\sin \frac{3\pi }{4}}=\frac{1-\left( -\frac{\sqrt{2}}{2} \right)}{\frac{\sqrt{2}}{2}}=\frac{1}{\frac{\sqrt{2}}{2}}+\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$ In order to solve the equation, take the individual L.C.M. $\tan \frac{3\pi }{8}=\frac{2}{\sqrt{2}}+1=\frac{2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}+1=\sqrt{2}+1$ And the rationalizing technique is used to compute the value in the aforementioned situation. Thus, the value of $\tan \frac{3\pi }{8}$ is $\sqrt{2}+1$.