Question
If
\[ \tan^{-1}\left(\frac{x+1}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \tan^{-1}(-7) \]
Find \( x \).
Solution
Use identity:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
Let
\[ a = \frac{x+1}{x-1}, \quad b = \frac{x-1}{x} \]
Compute numerator:
\[ a + b = \frac{x+1}{x-1} + \frac{x-1}{x} = \frac{x(x+1) + (x-1)^2}{x(x-1)} \]
\[ = \frac{x^2 + x + x^2 – 2x + 1}{x(x-1)} = \frac{2x^2 – x + 1}{x(x-1)} \]
Compute denominator:
\[ ab = \frac{x+1}{x-1} \cdot \frac{x-1}{x} = \frac{x+1}{x} \]
\[ 1 – ab = 1 – \frac{x+1}{x} = \frac{x – (x+1)}{x} = -\frac{1}{x} \]
Thus:
\[ \frac{a+b}{1-ab} = \frac{2x^2 – x + 1}{x(x-1)} \div \left(-\frac{1}{x}\right) \]
\[ = -\frac{2x^2 – x + 1}{x-1} \]
Given:
\[ -\frac{2x^2 – x + 1}{x-1} = -7 \]
\[ \frac{2x^2 – x + 1}{x-1} = 7 \]
\[ 2x^2 – x + 1 = 7x – 7 \]
\[ 2x^2 – 8x + 8 = 0 \]
\[ x^2 – 4x + 4 = 0 \Rightarrow (x – 2)^2 = 0 \]
\[ x = 2 \]
Final Answer:
\[ \boxed{2} \]
Key Concept
Use tangent addition identity and simplify algebra carefully.