Solving for tan(alpha) in Inverse Trigonometric Sum
The problem requires finding the value of $\tan \alpha$, where $\alpha$ is defined by the equation:
$ \alpha = \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) $We will simplify the summation term first.
Simplifying the Inverse Tangent Sum
Recall the inverse tangent subtraction identity:
$ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right) $We can rewrite the term inside the summation:
$ \frac{2^{p-1}}{1 + 2^{2p-1}} = \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} $Here, let $x = 2^p$ and $y = 2^{p-1}$. Applying the identity, the summation term becomes:
$ \tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \right) = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1}) $The summation is a telescoping series:
$ \sum_{p=1}^{11} (\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})) $Expanding the sum:
- $p=1: \tan^{-1}(2^1) - \tan^{-1}(2^0)$
- $p=2: \tan^{-1}(2^2) - \tan^{-1}(2^1)$
- ...
- $p=11: \tan^{-1}(2^{11}) - \tan^{-1}(2^{10})$
After cancellation, the sum simplifies to:
$ \tan^{-1}(2^{11}) - \tan^{-1}(2^0) = \tan^{-1}(2048) - \tan^{-1}(1) $Calculating alpha
Substitute the simplified sum back into the equation for $\alpha$:
$ \alpha = \frac{\pi}{4} + (\tan^{-1}(2048) - \tan^{-1}(1)) $Since $\frac{\pi}{4} = \tan^{-1}(1)$, we have:
$ \alpha = \tan^{-1}(1) + \tan^{-1}(2048) - \tan^{-1}(1) $ $ \alpha = \tan^{-1}(2048) $Finding tan(alpha)
The final step is to find the tangent of $\alpha$:
$ \tan \alpha = \tan(\tan^{-1}(2048)) $ $ \tan \alpha = 2048 $Similar Questions
- The number of solutions of $\tan^{-1} 4x + \tan^{-1} 6x = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
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Considering the principal values of inverse trigonometric functions, the value of the expression $\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
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Important Questions from Trigonometry
- The number of solutions of $\tan^{-1} 4x + \tan^{-1} 6x = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
Considering the principal values of inverse trigonometric functions, the value of the expression $\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right) - 2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
