This question was uploaded on 18/10/21 on social media accounts.
A, B, C are any three real numbers such that inverse tangents for these numbers are in A.P. And also AC = 1 + A + C. Find the value of B.
Solution:
Given that:
\[AC = 1 + A + C\]
\[\Rightarrow \frac{1-AC}{A+C} = -1\]
Now, arctan are in AP:
\[\Rightarrow \tan^{-1}A+\tan^{-1}C = 2\tan^{-1}B\]
\[\Rightarrow \tan^{-1}\left(\frac{A+C}{1-AC}\right) = \tan^{-1}\left(\frac{2B}{1-B^2}\right)\]
\[\Rightarrow \frac{A+C}{1-AC} = \frac{2B}{1-B^2}\]
\[\Rightarrow -1 = \frac{2B}{1-B^2}\]
\[\Rightarrow B^2-2B-1=0\]
\[\Rightarrow B=1\pm\sqrt2\]
Question related to Algebra, Trigonometry, Equation.
