The value of $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$ is

Question

Let $(a, b) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>, \theta \in(0,2 \pi)$, holds If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :

Answer 1

To solve the expression $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$, we can use trigonometric identities and properties of inverse functions.

The expression we are dealing with is $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$. We can simplify this using the tangent subtraction formula, which is: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. Here, $A = 2\tan^{-1}\left(\frac{1}{5}\right)$ and $B = \frac{\pi}{4}$.
First, we need $\tan A$. If $A = 2\tan^{-1}\left(\frac{1}{5}\right)$, use the identity: $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$. Let $\theta = \tan^{-1}(\frac{1}{5})$, then $\tan \theta = \frac{1}{5}$. Substituting, we have: $\tan(2\theta) = \frac{2 \times \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} = \frac{\frac{2}{5}}{1 - \frac{1}{25}} = \frac{\frac{2}{5}}{\frac{24}{25}}$. Simplifying: $\tan A = \frac{2}{5} \times \frac{25}{24} = \frac{10}{24} = \frac{5}{12}$.
Now let us incorporate the other part, $B = \frac{\pi}{4}$, and recall $\tan \frac{\pi}{4} = 1$.
Substitute these values into the subtraction formula: $\tan(A - B) = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \times 1} = \frac{\frac{5}{12} - \frac{12}{12}}{1 + \frac{5}{12}}$. Simplify the numerator: $\frac{5}{12} - \frac{12}{12} = -\frac{7}{12}$. Denominator becomes: $1 + \frac{5}{12} = \frac{12}{12} + \frac{5}{12} = \frac{17}{12}$.
Putting it all together: $\tan(A - B) = \frac{-\frac{7}{12}}{\frac{17}{12}} = -\frac{7}{12} \times \frac{12}{17} = -\frac{7}{17}$.

Therefore, the value of $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$ is $-\frac{7}{17}$.

Answer 2

To solve the expression $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$, we can use trigonometric identities and properties of inverse functions.

The expression we are dealing with is $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$. We can simplify this using the tangent subtraction formula, which is: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. Here, $A = 2\tan^{-1}\left(\frac{1}{5}\right)$ and $B = \frac{\pi}{4}$.
First, we need $\tan A$. If $A = 2\tan^{-1}\left(\frac{1}{5}\right)$, use the identity: $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$. Let $\theta = \tan^{-1}(\frac{1}{5})$, then $\tan \theta = \frac{1}{5}$. Substituting, we have: $\tan(2\theta) = \frac{2 \times \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} = \frac{\frac{2}{5}}{1 - \frac{1}{25}} = \frac{\frac{2}{5}}{\frac{24}{25}}$. Simplifying: $\tan A = \frac{2}{5} \times \frac{25}{24} = \frac{10}{24} = \frac{5}{12}$.
Now let us incorporate the other part, $B = \frac{\pi}{4}$, and recall $\tan \frac{\pi}{4} = 1$.
Substitute these values into the subtraction formula: $\tan(A - B) = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \times 1} = \frac{\frac{5}{12} - \frac{12}{12}}{1 + \frac{5}{12}}$. Simplify the numerator: $\frac{5}{12} - \frac{12}{12} = -\frac{7}{12}$. Denominator becomes: $1 + \frac{5}{12} = \frac{12}{12} + \frac{5}{12} = \frac{17}{12}$.
Putting it all together: $\tan(A - B) = \frac{-\frac{7}{12}}{\frac{17}{12}} = -\frac{7}{12} \times \frac{12}{17} = -\frac{7}{17}$.

Therefore, the value of $\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]$ is $-\frac{7}{17}$.

The value of \(\tan\left[2\tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right]\) is

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The Correct Option is D

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