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)}$Multiply numerator and denominator by $(2^x + 1)(1 + 2^{x+1})$:$	an(\alpha + \beta) = \frac{2^x(1 + 2^{x+1}) + (2^x + 1)}{(2^x + 1)(1 + 2^{x+1}) - 2^x}$$= \frac{2^x + 2^x \cdot 2^{x+1} + 2^x + 1}{2^x + 2^x \cdot 2^{x+1} + 1 + 2^{x+1} - 2^x}$$= \frac{2 \cdot 2^x + 2 \cdot 2^{2x} + 1}{2 \cdot 2^{2x} + 2^{x+1} + 1}$Since $2^{x+1} = 2 \cdot 2^x$,the numerator is $2 \cdot 2^x + 2 \cdot 2^{2x} + 1$ and the denominator is $2 \cdot 2^{2x} + 2 \cdot 2^x + 1$.Thus,$	an(\alpha + \beta) = 1$.Therefore,$\alpha + \beta = \frac{\pi}{4}$.

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

Similar Questions

The value of $\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$ is equal to

If $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$,where $A$ and $B$ are positive acute angles,then $A + B = $

Prove that $\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)$

$\frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module