Question
If
\[ \tan^{-1}3 + \tan^{-1}x = \tan^{-1}8 \]
Find \( x \).
Solution
Use identity:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
So,
\[ \tan^{-1}\left(\frac{3 + x}{1 – 3x}\right) = \tan^{-1}(8) \]
Thus,
\[ \frac{3 + x}{1 – 3x} = 8 \]
\[ 3 + x = 8 – 24x \]
\[ 25x = 5 \Rightarrow x = \frac{1}{5} \]
Final Answer:
\[ \boxed{\frac{1}{5}} \]
Key Concept
Apply tangent addition identity and equate arguments carefully.