Ts Eamcet
Inverse Trigonometric Functions
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ+1 / -02020
If $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$, then $x^2=$
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