What is \(\tan ^{- 1}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{3}{5}} \right)\) equal to?
π/4
Understanding the Problem: Finding the Sum of Inverse Tangents
The question asks us to find the value of the expression \( \tan^{-1}\left( \frac{1}{4} \right) + \tan^{-1}\left( \frac{3}{5} \right) \). This involves adding two inverse tangent functions. To solve this, we will use a standard formula for the sum of two inverse tangent functions.
Key Formula for Sum of Inverse Tangents
The primary formula used to add two inverse tangent values is:
\(\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)\), provided that \(xy < 1\).
There are other versions of this formula depending on the values of \(x\), \(y\), and \(xy\), but the condition \(xy < 1\) covers the most common cases and applies here.
Step-by-Step Calculation
Let's identify \(x\) and \(y\) from the given expression:
- \(x = \frac{1}{4}\)
- \(y = \frac{3}{5}\)
First, we must check the condition \(xy < 1\):
\(xy = \left(\frac{1}{4}\right) \times \left(\frac{3}{5}\right) = \frac{1 \times 3}{4 \times 5} = \frac{3}{20}\)
Since \( \frac{3}{20} < 1 \), the condition \(xy < 1\) is satisfied, and we can use the formula \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).
Now, let's calculate the numerator and the denominator of the fraction inside the \(\tan^{-1}\) function:
Numerator (\(x+y\)):
\(x+y = \frac{1}{4} + \frac{3}{5}\)
To add these fractions, we find a common denominator, which is 20.
\(x+y = \frac{1 \times 5}{4 \times 5} + \frac{3 \times 4}{5 \times 4} = \frac{5}{20} + \frac{12}{20} = \frac{5+12}{20} = \frac{17}{20}\)
Denominator (\(1-xy\)):
\(1-xy = 1 - \frac{3}{20}\)
To subtract, we write 1 as \(\frac{20}{20}\).
\(1-xy = \frac{20}{20} - \frac{3}{20} = \frac{20-3}{20} = \frac{17}{20}\)
Now, substitute these values back into the formula:
\(\tan^{-1}\left(\frac{x+y}{1-xy}\right) = \tan^{-1}\left(\frac{\frac{17}{20}}{\frac{17}{20}}\right)\)
Simplify the fraction inside the \(\tan^{-1}\) function:
\(\frac{\frac{17}{20}}{\frac{17}{20}} = 1\)
So, the expression simplifies to:
\(\tan^{-1}(1)\)
Finally, we need to find the value of \( \tan^{-1}(1) \). The inverse tangent of 1 is the angle whose tangent is 1. This angle in the principal value range \((-\frac{\pi}{2}, \frac{\pi}{2})\) is \( \frac{\pi}{4} \).
\(\tan^{-1}(1) = \frac{\pi}{4}\)
Therefore, \( \tan^{-1}\left( \frac{1}{4} \right) + \tan^{-1}\left( \frac{3}{5} \right) = \frac{\pi}{4} \).
Comparing with Options
Let's check our result against the given options:
- Option 1: 0
- Option 2: \(\pi/4\)
- Option 3: \(\pi/3\)
- Option 4: \(\pi/2\)
Our calculated value, \( \frac{\pi}{4} \), matches Option 2.
| Expression | Calculated Value |
|---|---|
| \(\tan^{-1}\left( \frac{1}{4} \right) + \tan^{-1}\left( \frac{3}{5} \right)\) | \(\frac{\pi}{4}\) |
Conclusion
Using the sum formula for inverse tangents, we found that the value of \( \tan^{-1}\left( \frac{1}{4} \right) + \tan^{-1}\left( \frac{3}{5} \right) \) is \( \frac{\pi}{4} \).
Revision Table: Inverse Trigonometric Functions
| Function | Domain | Principal Value Range | Key Values (Examples) |
|---|---|---|---|
| \(\sin^{-1}(x)\) | [-1, 1] | \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) | \(\sin^{-1}(0)=0\), \(\sin^{-1}(1)=\frac{\pi}{2}\) |
| \(\cos^{-1}(x)\) | [-1, 1] | \( [0, \pi] \) | \(\cos^{-1}(0)=\frac{\pi}{2}\), \(\cos^{-1}(1)=0\) |
| \(\tan^{-1}(x)\) | \( (-\infty, \infty) \) | \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) | \(\tan^{-1}(0)=0\), \(\tan^{-1}(1)=\frac{\pi}{4}\) |
Additional Information: Properties of Inverse Tangent
The inverse tangent function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is the inverse of the tangent function. It gives the angle whose tangent is \(x\).
- Domain: The domain of \( \tan^{-1}(x) \) is all real numbers, \( (-\infty, \infty) \), because the range of \( \tan(x) \) is \( (-\infty, \infty) \).
- Range (Principal Value): To make the inverse function unique, we restrict the range of \( \tan(x) \) to \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Therefore, the principal value range of \( \tan^{-1}(x) \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
- Graph: The graph of \( y = \tan^{-1}(x) \) has horizontal asymptotes at \( y = \frac{\pi}{2} \) and \( y = -\frac{\pi}{2} \). It is an increasing function.
- Other Important Formulas:
- \( \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \), provided \( xy > -1 \).
- \( 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), provided \( |x| < 1 \).
Understanding these properties and formulas is crucial for solving problems involving inverse trigonometric functions, especially in calculus and trigonometry.
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