If tan alpha = frac{x^2 - x}{x^2 - x + 1} and | vedclass

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

Question

(A) Let $t = x^2 - x$. Then $	an \alpha = \frac{t}{t + 1}$ and $	an \beta = \frac{1}{2t + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $t = x^2 - x$. Then $	an \alpha = \frac{t}{t + 1}$ and $	an \beta = \frac{1}{2t + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we substitute the values:$	an(\alpha + \beta) = \frac{\frac{t}{t + 1} + \frac{1}{2t + 1}}{1 - \left(\frac{t}{t + 1}ight) \left(\frac{1}{2t + 1}ight)}$$= \frac{\frac{t(2t + 1) + (t + 1)}{(t + 1)(2t + 1)}}{\frac{(t + 1)(2t + 1) - t}{(t + 1)(2t + 1)}}$$= \frac{2t^2 + t + t + 1}{2t^2 + 2t + t + 1 - t}$$= \frac{2t^2 + 2t + 1}{2t^2 + 2t + 1} = 1$Therefore,$	an(\alpha + \beta) = 1$.

If $\tan \alpha = \frac{x^2 - x}{x^2 - x + 1}$ and $\tan \beta = \frac{1}{2x^2 - 2x + 1}$ $(x \ne 0, 1)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\tan(\alpha + \beta)$ has the value equal to:

Similar Questions

$16 \sin(20^{\circ}) \sin(40^{\circ}) \sin(80^{\circ})$ is equal to

If $\cosh \beta = \sec \alpha \cos \theta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin \theta$,then $\sinh^2 \beta =$

All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot \frac{1}{4^{\sin^2 y}} \leq 1$ also satisfy the equation

If $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15} = x$,then $\frac{1}{8x} =$

$\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module