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CUET
-- Mathematics - Section B1
Inverse Trigonometric Functions
The value of $tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}$, is
0
$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{4}$
We have
$tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}= tan^{-1}\left(\frac{1/2+1/3}{1-1/2×1/3}\right) = tan^{-1}1=\frac{\pi}{4}$