The value of tan frac{pi}{8} is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$.Setting $	heta = \frac{\pi}{8}$,we have $	an \frac{\pi}{4} = \frac{2	an \fra

Vedclass · Accepted Answer

$\sqrt{2}-1$

Vedclass · Answer

(C) We know that $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$.Setting $	heta = \frac{\pi}{8}$,we have $	an \frac{\pi}{4} = \frac{2	an \frac{\pi}{8}}{1-	an^2 \frac{\pi}{8}}$.Since $	an \frac{\pi}{4} = 1$,let $y = 	an \frac{\pi}{8}$.Then $1 = \frac{2y}{1-y^2}$,which implies $1 - y^2 = 2y$,or $y^2 + 2y - 1 = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$.Since $\frac{\pi}{8}$ is in the $1^{	ext{st}}$ quadrant,$	an \frac{\pi}{8} > 0$.Therefore,$	an \frac{\pi}{8} = \sqrt{2} - 1$.

The value of $\tan \frac{\pi}{8}$ is

Similar Questions

$\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta} = $

If $\theta$ does not lie in the second quadrant and $\tan \theta = \frac{-3}{4}$,then $\tan \frac{\theta}{2} + \sin 2 \theta =$

If $\tan \alpha = \frac{-12}{5}$,$\cot \beta = \frac{7}{24}$,$\alpha$ does not belong to the second quadrant,and $\beta$ does not belong to the first quadrant,then $\sqrt{13} \sin \frac{\alpha}{2} + \cos \frac{\beta}{2} + \tan \frac{\alpha}{2} \cot \frac{\beta}{2} = $

If $\sin \alpha = \frac{-3}{5}$ where $\pi < \alpha < \frac{3\pi}{2},$ then $\cos \frac{\alpha}{2} = $

If $\tan \theta = \frac{1}{3}$,then $\cos 2 \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module