If tan alpha = frac{m}{m + 1} and tan beta = | vedclass

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

Question

(B) We have $	an \alpha = \frac{m}{m + 1}$ and $	an \beta = \frac{1}{2m + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \be

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) We have $	an \alpha = \frac{m}{m + 1}$ and $	an \beta = \frac{1}{2m + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$:$	an(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \left(\frac{m}{m + 1}ight) \left(\frac{1}{2m + 1}ight)}$$= \frac{\frac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1)}}{\frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)}}$$= \frac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}$$= \frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$Since $	an(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

If $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$,then $\alpha + \beta = $

Similar Questions

If $\cos (\alpha+\beta)=\frac{4}{5}$,$\sin (\alpha-\beta)=\frac{5}{13}$ and $\alpha, \beta$ are between $0$ and $\frac{\pi}{4}$,then $\tan 2 \alpha$ is equal to

$\frac{\sinh(x+y) + \sinh(x-y)}{\cosh(x+y) - \cosh(x-y)} = $

The value of $\frac{\tan 70^o - \tan 20^o}{\tan 50^o} = $

Prove that $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$

If $\tan A = \frac{1}{\sqrt{x(x^2+x+1)}}, \tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C = (x^{-3}+x^{-2}+x^{-1})^{\frac{1}{2}}$,where $0 < A, B, C < \frac{\pi}{2}$,then $A+B$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module