If tan alpha = (1 + 2^{-x})^{-1} and tan beta | vedclass

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

Question

(B) Given $	an \alpha = \frac{1}{1 + 2^{-x}} = \frac{2^x}{2^x + 1}$ and $	an \beta = \frac{1}{1 + 2^{x+1}}$.Using the formula $	an(\alpha + \beta)

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(B) Given $	an \alpha = \frac{1}{1 + 2^{-x}} = \frac{2^x}{2^x + 1}$ and $	an \beta = \frac{1}{1 + 2^{x+1}}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$:$	an(\alpha + \beta) = \frac{\frac{2^x}{2^x + 1} + \frac{1}{1 + 2^{x+1}}}{1 - \left(\frac{2^x}{2^x + 1}ight) \left(\frac{1}{1 + 2^{x+1}}ight)}$Simplify the numerator: $\frac{2^x(1 + 2^{x+1}) + (2^x + 1)}{(2^x + 1)(1 + 2^{x+1})} = \frac{2^x + 2^{2x+1} + 2^x + 1}{(2^x + 1)(1 + 2^{x+1})} = \frac{2 \cdot 2^x + 2 \cdot 2^{2x} + 1}{(2^x + 1)(1 + 2^{x+1})}$.Simplify the denominator: $\frac{(2^x + 1)(1 + 2^{x+1}) - 2^x}{(2^x + 1)(1 + 2^{x+1})} = \frac{2^x + 2^{2x+1} + 1 + 2^{x+1} - 2^x}{(2^x + 1)(1 + 2^{x+1})} = \frac{2^{2x+1} + 2^{x+1} + 1}{(2^x + 1)(1 + 2^{x+1})}$.Since the numerator and denominator are equal, $	an(\alpha + \beta) = 1$.Therefore, $\alpha + \beta = \pi / 4$.

If $\tan \alpha = (1 + 2^{-x})^{-1}$ and $\tan \beta = (1 + 2^{x+1})^{-1}$, then $\alpha + \beta$ equals:

Similar Questions

If $x + \frac{1}{x} = 2\cos \theta ,$ then ${x^3} + \frac{1}{{{x^3}}} = $

If $a \tan \theta = b$,then $a \cos 2\theta + b \sin 2\theta = $

$\frac{\cos A}{1 - \sin A} = $

If $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$,then $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ must be equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module