The value of \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is
We need to find the value of the given trigonometric expression: \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\). This expression involves the tangent function applied to a difference of two angles, one of which is related to an inverse tangent function and the other is a standard angle.
Let the expression be denoted by \(E\). We can write it in the form \(E = \tan(A - B)\), where \(A = 2{{\tan }^{ - 1}}\frac{1}{5}\) and \(B = \frac{\pi }{4}\).
To evaluate this, we will use two important trigonometric identities:
Let's first simplify \(A = 2{{\tan }^{ - 1}}\frac{1}{5}\). Here, \(x = \frac{1}{5}\). Since \(|\frac{1}{5}| = \frac{1}{5} < 1\), we can use the identity \(2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\).
Substituting \(x = \frac{1}{5}\):
\[2{{\tan }^{ - 1}}\frac{1}{5} = {{\tan }^{ - 1}}\frac{{2 \times \frac{1}{5}}}{{1 - {{\left(\frac{1}{5}\right)}^2}}}\] \[ = {{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}\] \[ = {{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{\frac{{25 - 1}}{{25}}}}\] \[ = {{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{\frac{{24}}{{25}}}}\] \[ = {{\tan }^{ - 1}}\left(\frac{2}{5} \times \frac{25}{24}\right)\] \[ = {{\tan }^{ - 1}}\left(\frac{2 \times 25}{5 \times 24}\right)\] \[ = {{\tan }^{ - 1}}\left(\frac{50}{120}\right)\] \[ = {{\tan }^{ - 1}}\frac{5}{12}\]So, \(A = {{\tan }^{ - 1}}\frac{5}{12}\).
We know that \(\tan\left(\frac{\pi}{4}\right) = 1\). Therefore, \(B = \frac{\pi}{4}\) can be written as \({{\tan }^{ - 1}}1\).
So, the original expression becomes \(E = {\rm{tan}}\left( {{{\tan }^{ - 1}}\frac{5}{12} - {{\tan }^{ - 1}}1} \right)\).
Now we have the expression in the form \({\rm{tan}}(A - B)\), where \(A = {{\tan }^{ - 1}}\frac{5}{12}\) and \(B = {{\tan }^{ - 1}}1\). Using the identity \({\rm{tan}}(X - Y) = \frac{{{\rm{tan}}X - {\rm{tan}}Y}}{{1 + {\rm{tan}}X{\rm{tan}}Y}}\):
Let \(X = {{\tan }^{ - 1}}\frac{5}{12}\) and \(Y = {{\tan }^{ - 1}}1\). Then \(\tan X = \frac{5}{12}\) and \(\tan Y = 1\).
\[E = \frac{{\tan \left({{\tan }^{ - 1}}\frac{5}{12}\right) - \tan \left({{\tan }^{ - 1}}1\right)}}{{1 + \tan \left({{\tan }^{ - 1}}\frac{5}{12}\right) \times \tan \left({{\tan }^{ - 1}}1\right)}}\] \[E = \frac{{\frac{5}{12} - 1}}{{1 + \frac{5}{12} \times 1}}\] \[E = \frac{{\frac{5 - 12}{12}}}{{1 + \frac{5}{12}}}\] \[E = \frac{{\frac{-7}{12}}}{{\frac{12 + 5}{12}}}\] \[E = \frac{{\frac{-7}{12}}}{{\frac{17}{12}}}\] \[E = \frac{-7}{12} \times \frac{12}{17}\] \[E = \frac{-7}{17}\]The value of the expression \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is \(-\frac{7}{17}\).
Here is a quick summary of the steps taken:
| Identity Type | Formula | Notes |
|---|---|---|
| Tangent Difference | \({\rm{tan}}(X - Y) = \frac{{{\rm{tan}}X - {\rm{tan}}Y}}{{1 + {\rm{tan}}X{\rm{tan}}Y}}\) | Used for simplifying \(\tan(A-B)\) form. |
| Inverse Tangent (Double Angle) | \(2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\) | Valid for \(|x| < 1\). |
| Inverse Tangent Property | \(\tan({{\tan }^{ - 1}}x) = x\) | Used when applying tangent function to its inverse. |
The inverse tangent function, \({{\tan }^{ - 1}}x\) or \(\arctan x\), has a specific domain and range that are important to remember when working with inverse trigonometric functions:
In our calculation, \(2{{\tan }^{ - 1}}\frac{1}{5} = {{\tan }^{ - 1}}\frac{5}{12}\). The value \(\frac{5}{12}\) is positive, so \({{\tan }^{ - 1}}\frac{5}{12}\) is in the range \(\left(0, \frac{\pi}{2}\right)\). The angle \(\frac{\pi}{4}\) is also in this range. When we subtract \(\frac{\pi}{4}\) from \({{\tan }^{ - 1}}\frac{5}{12}\), the result could be positive or negative, depending on which angle is larger. Since \(\frac{5}{12} < 1\), \({{\tan }^{ - 1}}\frac{5}{12} < {{\tan }^{ - 1}}1 = \frac{\pi}{4}\). Therefore, \({{\tan }^{ - 1}}\frac{5}{12} - \frac{\pi}{4}\) will be a negative angle, and its tangent will be negative, which aligns with our result \(-\frac{7}{17}\).
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