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

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

Question

(C) We use the formula for the tangent of a sum of two angles: $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Su

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) We use the formula for the tangent of a sum of two angles: $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Substituting the given values:$	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)}$Simplifying the numerator: $\frac{m(2m+1) + 1(m+1)}{(m+1)(2m+1)} = \frac{2m^2 + m + m + 1}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}$.Simplifying the denominator: $1 - \frac{m}{(m+1)(2m+1)} = \frac{(m+1)(2m+1) - m}{(m+1)(2m+1)} = \frac{2m^2 + 3m + 1 - m}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}$.Dividing the numerator by the denominator:$	an(\alpha + \beta) = \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 $\sin 3A = \cos (A - 26^{\circ})$,where $3A$ is an acute angle,then the value of $A$ is (in $^{\circ}$)

If the roots of the quadratic equation $x^2 + px + q = 0$ are $\tan 30^\circ$ and $\tan 15^\circ$ respectively,then the value of $2 + q - p$ is

The value of $x$ for the maximum value of $\sqrt{3} \cos x + \sin x$ is .....$^o$

If $A$ lies in the third quadrant and $3\tan A - 4 = 0,$ then $5\sin 2A + 3\sin A + 4\cos A = $

The minimum value of $3 \cos x + 4 \sin x + 5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module