If \(\tan^{-1} \left(\frac{1}{2}\right)+\tan^{-1} \left(\frac{x}{3}\right)=\frac{\pi}{4},\) where 0 < x < 6, then what is x equal to?
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Solving Inverse Trigonometric Equations
We are asked to find the value of \(x\) that satisfies the equation \(\tan^{-1} \left(\frac{1}{2}\right)+\tan^{-1} \left(\frac{x}{3}\right)=\frac{\pi}{4}\), given the condition \(0 < x < 6\).
To solve this inverse trigonometric equation, we can use the sum formula for inverse tangents: \(\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)\), which is valid when \(AB < 1\).
In our equation, we have \(A = \frac{1}{2}\) and \(B = \frac{x}{3}\).
First, let's check the condition \(AB < 1\):
\(AB = \left(\frac{1}{2}\right) \cdot \left(\frac{x}{3}\right) = \frac{x}{6}\)
Since the problem states that \(0 < x < 6\), it implies that \(\frac{0}{6} < \frac{x}{6} < \frac{6}{6}\), which means \(0 < \frac{x}{6} < 1\). Thus, the condition \(AB < 1\) is satisfied for the given range of \(x\), and we can safely use the sum formula.
Applying the Tan Inverse Sum Formula
Using the formula \(\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)\), we transform the left side of the equation:
\(\tan^{-1} \left(\frac{1}{2}\right)+\tan^{-1} \left(\frac{x}{3}\right) = \tan^{-1} \left(\frac{\frac{1}{2}+\frac{x}{3}}{1-\left(\frac{1}{2}\right)\left(\frac{x}{3}\right)}\right)\)
Simplify the expression inside the \(\tan^{-1}\):
Numerator: \(\frac{1}{2}+\frac{x}{3} = \frac{3}{6}+\frac{2x}{6} = \frac{3+2x}{6}\)
Denominator: \(1-\frac{x}{6} = \frac{6}{6}-\frac{x}{6} = \frac{6-x}{6}\)
So, the expression becomes:
\(\frac{\frac{3+2x}{6}}{\frac{6-x}{6}} = \frac{3+2x}{6} \cdot \frac{6}{6-x} = \frac{3+2x}{6-x}\)
The original equation now becomes:
\(\tan^{-1} \left(\frac{3+2x}{6-x}\right) = \frac{\pi}{4}\)
Solving for x
To find \(x\), we can take the tangent of both sides of the equation:
\(\tan \left(\tan^{-1} \left(\frac{3+2x}{6-x}\right)\right) = \tan \left(\frac{\pi}{4}\right)\)
We know that \(\tan(\tan^{-1} y) = y\) and \(\tan\left(\frac{\pi}{4}\right) = 1\). So, the equation simplifies to:
\(\frac{3+2x}{6-x} = 1\)
Now, we solve this algebraic equation for \(x\). Multiply both sides by \((6-x)\):
\(3+2x = 1 \cdot (6-x)\)
\(3+2x = 6-x\)
Gather the terms with \(x\) on one side and constant terms on the other:
\(2x + x = 6 - 3\)
\(3x = 3\)
Divide by 3:
\(x = \frac{3}{3}\)
\(x = 1\)
Verification
We must verify if our solution \(x=1\) satisfies the given condition \(0 < x < 6\). Since \(0 < 1 < 6\) is true, the value \(x=1\) is a valid solution.
Thus, the value of \(x\) is 1.
Comparing with Options
The calculated value of \(x=1\) matches option 1.
| Step | Process | Result |
|---|---|---|
| 1 | Identify the given equation and formula | \(\tan^{-1} \left(\frac{1}{2}\right)+\tan^{-1} \left(\frac{x}{3}\right)=\frac{\pi}{4}\), Use \(\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)\) |
| 2 | Check the condition \(AB < 1\) | \(A=\frac{1}{2}, B=\frac{x}{3}\). \(AB = \frac{x}{6}\). Since \(0 < x < 6\), \(0 < \frac{x}{6} < 1\). Condition satisfied. |
| 3 | Apply the sum formula | \(\tan^{-1} \left(\frac{\frac{1}{2}+\frac{x}{3}}{1-\frac{x}{6}}\right) = \frac{\pi}{4}\) |
| 4 | Simplify the argument | \(\tan^{-1} \left(\frac{3+2x}{6-x}\right) = \frac{\pi}{4}\) |
| 5 | Take tangent of both sides | \(\frac{3+2x}{6-x} = \tan\left(\frac{\pi}{4}\right) = 1\) |
| 6 | Solve the resulting equation | \(\frac{3+2x}{6-x} = 1 \implies 3+2x = 6-x \implies 3x = 3 \implies x=1\) |
| 7 | Verify the solution | \(x=1\) satisfies \(0 < x < 6\). |
Revision Table: Inverse Trigonometric Functions
| Function | Domain | Range |
|---|---|---|
| \(\sin^{-1} x\) | \([-1, 1]\) | \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) |
| \(\cos^{-1} x\) | \([-1, 1]\) | \([0, \pi]\) |
| \(\tan^{-1} x\) | \((-\infty, \infty)\) | \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) |
| \(\cot^{-1} x\) | \((-\infty, \infty)\) | \((0, \pi)\) |
| \(\sec^{-1} x\) | \((-\infty, -1] \cup [1, \infty)\) | \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\) |
| \(\csc^{-1} x\) | \((-\infty, -1] \cup [1, \infty)\) | \([-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]\) |
Additional Information on Tan Inverse Formulas
The formula \(\tan^{-1} A + \tan^{-1} B\) has different variations depending on the values of A and B:
- If \(AB < 1\), then \(\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)\)
- If \(AB > 1\), and \(A, B > 0\), then \(\tan^{-1} A + \tan^{-1} B = \pi + \tan^{-1} \left(\frac{A+B}{1-AB}\right)\)
- If \(AB > 1\), and \(A, B < 0\), then \(\tan^{-1} A + \tan^{-1} B = -\pi + \tan^{-1} \left(\frac{A+B}{1-AB}\right)\)
- If \(AB = 1\), and \(A, B > 0\), then \(\tan^{-1} A + \tan^{-1} B = \frac{\pi}{2}\)
- If \(AB = 1\), and \(A, B < 0\), then \(\tan^{-1} A + \tan^{-1} B = -\frac{\pi}{2}\)
It is crucial to check the condition on \(AB\) before applying the sum formula to avoid errors.
Important Questions from Inverse Trigonometric Functions
What is \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) equal to ?
What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?
Consider the following statements:
1. There exists \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) for which tan -1 (tan θ) ≠ θ
2. \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{15}}} \right)\)
Which of the above statements is/are correct?
Consider the following statements:
1. \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\)
2. There exist x, y ∈ [-1, 1], where x ≠ y such that sin -1 x + cos -1 \({\rm{y}} = \frac{{\rm{\pi }}}{2}\)
Which of the above statements is/are correct?The value of \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is
