The value of tan left(frac{7 pi}{8}right) is | vedclass

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

Question

(B) Let $A = 	an \left(\frac{7 \pi}{8}ight)$.Since $\frac{7 \pi}{8} = \pi - \frac{\pi}{8}$,we have $A = 	an \left(\pi - \frac{\pi}{8}ight) = -

Vedclass · Accepted Answer

$1-\sqrt{2}$

Vedclass · Answer

(B) Let $A = 	an \left(\frac{7 \pi}{8}ight)$.Since $\frac{7 \pi}{8} = \pi - \frac{\pi}{8}$,we have $A = 	an \left(\pi - \frac{\pi}{8}ight) = -	an \left(\frac{\pi}{8}ight)$.Using the formula $	an(2	heta) = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,let $	heta = \frac{\pi}{8}$.Then $	an \left(\frac{\pi}{4}ight) = \frac{2 	an \left(\frac{\pi}{8}ight)}{1 - 	an^2 \left(\frac{\pi}{8}ight)} = 1$.Substituting $	an \left(\frac{\pi}{8}ight) = -A$,we get $\frac{2(-A)}{1 - (-A)^2} = 1$.$-2A = 1 - A^2 \Rightarrow A^2 - 2A - 1 = 0$.Using the quadratic formula $A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $A = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.Since $\frac{7 \pi}{8}$ lies in the $2^{nd}$ quadrant,$	an \left(\frac{7 \pi}{8}ight) Thus,$A = 1 - \sqrt{2}$.

The value of $\tan \left(\frac{7 \pi}{8}\right)$ is

Similar Questions

The value of $x$ in $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $(\sin x)(\cos x) = \frac{1}{4}$ is

If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $

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

The value of $\cos 15^{\circ} \cos 7.5^{\circ} \sin 7.5^{\circ}$ is

Evaluate: $\cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module