Mathematics · 2022
JEE Main · 26 July 2022, Shift 1 · Q19
tan (2 tan^-1 1/5+ sec^-1 (√ 5)/2+2 tan^-1 1/8) is equal to:
$\displaystyle \tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$ is equal to :
Official answer
(2)
$\displaystyle 2$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.