The equation \({\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\) is satisfied by
x = 0
Solving Inverse Tangent Equation for x
We are asked to find the value of \({\rm{x}}\) that satisfies the equation:
\[{\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\]
This equation involves the sum of two inverse tangent functions. We can use properties of inverse trigonometric functions to solve this.
Understanding the Key Property
One important property of inverse tangent functions is:
- If \(ab = 1\), then \({\tan ^{ - 1}}(a) + {\tan ^{ - 1}}(b) = \frac{\pi}{2}\) (assuming \(a > 0, b > 0\)).
- If \(ab < 1\), then \({\tan ^{ - 1}}(a) + {\tan ^{ - 1}}(b) = {\tan ^{ - 1}}\left(\frac{a+b}{1-ab}\right)\).
- If \(ab > 1\), then \({\tan ^{ - 1}}(a) + {\tan ^{ - 1}}(b) = \pi + {\tan ^{ - 1}}\left(\frac{a+b}{1-ab}\right)\) (if \(a > 0, b > 0\)).
Given that the sum is equal to \(\frac{\pi}{2}\), the property \({\tan ^{ - 1}}(a) + {\tan ^{ - 1}}(b) = \frac{\pi}{2}\) when \(ab=1\) is highly relevant here. In our equation, \(a = 1+x\) and \(b = 1-x\).
Applying the Property to Solve for x
For the equation \({\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\) to hold, we can assume that the product of the arguments \( (1+x) \) and \( (1-x) \) must be equal to 1. Let's set up the equation based on this property:
\[(1 + {\rm{x}})(1 - {\rm{x}}) = 1\]
This is a difference of squares expression on the left side, which simplifies as \((a+b)(a-b) = a^2 - b^2\).
\[1^2 - {\rm{x}}^2 = 1\]
\[1 - {\rm{x}}^2 = 1\]
Now, we solve for \({\rm{x}}\):
Subtract 1 from both sides:
\[ - {\rm{x}}^2 = 1 - 1\]
\[ - {\rm{x}}^2 = 0\]
Multiply by -1:
\[{\rm{x}}^2 = 0\]
Taking the square root of both sides:
\[{\rm{x}} = 0\]
Verification
Let's check if \({\rm{x}} = 0\) satisfies the original equation:
Substitute \({\rm{x}} = 0\) into the equation \({\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\):
\[{\tan ^{ - 1}}\left( {1 + 0} \right) + {\tan ^{ - 1}}\left( {1 - 0} \right)\]
\[{\tan ^{ - 1}}\left( 1 \right) + {\tan ^{ - 1}}\left( 1 \right)\]
We know that \({\tan ^{ - 1}}(1) = \frac{{\rm{\pi }}}{4}\).
\[\frac{{\rm{\pi }}}{4} + \frac{{\rm{\pi }}}{4} = \frac{{2{\rm{\pi }}}}{4} = \frac{{\rm{\pi }}}{2}\]
The left side equals the right side, so \({\rm{x}} = 0\) is indeed the correct solution.
The condition \(ab=1\) requires that both \(a\) and \(b\) have the same sign. When \(x=0\), \(a=1+0=1\) and \(b=1-0=1\). Both are positive, satisfying the typical condition for the \(ab=1 \implies \frac{\pi}{2}\) rule when the arguments are positive.
Conclusion
The value of \({\rm{x}}\) that satisfies the given equation is \({\rm{x}} = 0\).
| Option | Value of x | Equation Check | Result |
|---|---|---|---|
| 1 | x = 1 | \({\tan ^{ - 1}}(1+1) + {\tan ^{ - 1}}(1-1) = {\tan ^{ - 1}}(2) + {\tan ^{ - 1}}(0) = {\tan ^{ - 1}}(2) + 0 = {\tan ^{ - 1}}(2)\) | \(\neq \frac{{\rm{\pi }}}{2}\) |
| 2 | x = -1 | \({\tan ^{ - 1}}(1-1) + {\tan ^{ - 1}}(1-(-1)) = {\tan ^{ - 1}}(0) + {\tan ^{ - 1}}(2) = 0 + {\tan ^{ - 1}}(2) = {\tan ^{ - 1}}(2)\) | \(\neq \frac{{\rm{\pi }}}{2}\) |
| 3 | x = 0 | \({\tan ^{ - 1}}(1+0) + {\tan ^{ - 1}}(1-0) = {\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(1) = \frac{{\rm{\pi }}}{4} + \frac{{\rm{\pi }}}{4} = \frac{{\rm{\pi }}}{2}\) | \(= \frac{{\rm{\pi }}}{2}\) |
| 4 | x = ½ | \({\tan ^{ - 1}}(1+\frac{1}{2}) + {\tan ^{ - 1}}(1-\frac{1}{2}) = {\tan ^{ - 1}}(\frac{3}{2}) + {\tan ^{ - 1}}(\frac{1}{2})\). Here \(ab = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} < 1\). Sum is \({\tan ^{ - 1}}\left(\frac{\frac{3}{2}+\frac{1}{2}}{1-\frac{3}{4}}\right) = {\tan ^{ - 1}}\left(\frac{2}{\frac{1}{4}}\right) = {\tan ^{ - 1}}(8)\). | \(\neq \frac{{\rm{\pi }}}{2}\) |
The correct option is the one stating \({\rm{x}} = 0\).
Revision Table: Inverse Trigonometric Properties
| Property | Condition | Formula |
|---|---|---|
| Sum of tan inverse | \(ab < 1\) | \({\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left(\frac{a+b}{1-ab}\right)\) |
| Sum of tan inverse | \(ab = 1, a > 0, b > 0\) | \({\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \frac{\pi}{2}\) |
| Sum of tan inverse | \(ab > 1, a > 0, b > 0\) | \({\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \pi + {\tan ^{ - 1}}\left(\frac{a+b}{1-ab}\right)\) |
| Sum of tan inverse | \(ab > 1, a < 0, b < 0\) | \({\tan ^{ - 1}}a + {\tan ^{ - 1}}b = -\pi + {\tan ^{ - 1}}\left(\frac{a+b}{1-ab}\right)\) |
| Inverse Relation | \({\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi}{2}\) |
Additional Information on Inverse Tangent Functions
Inverse tangent, denoted as \({\tan ^{ - 1}}({\rm{x}})\) or \(\arctan({\rm{x}})\), is the inverse function of the tangent function. It gives the angle whose tangent is \({\rm{x}}\).
- The domain of \({\tan ^{ - 1}}({\rm{x}})\) is all real numbers, \((-\infty, \infty)\).
- The principal value range of \({\tan ^{ - 1}}({\rm{x}})\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\). This means the output angle is always between -\(\frac{\pi}{2}\) and \(\frac{\pi}{2}\) (exclusive of the endpoints).
- The graph of \(y = {\tan ^{ - 1}}({\rm{x}})\) has horizontal asymptotes at \(y = \frac{\pi}{2}\) and \(y = -\frac{\pi}{2}\).
- \({\tan ^{ - 1}}({\rm{x}})\) is an odd function, meaning \({\tan ^{ - 1}}(-{\rm{x}}) = -{\tan ^{ - 1}}({\rm{x}})\).
Understanding these properties is crucial for solving equations involving inverse trigonometric functions.
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