The value of (\tan^{-1}(\dfrac{1}{2})+\tan^{-1}(\dfrac{2}{5}... | Mathematics | SolveeIt

The value of $\tan^{-1}(\dfrac{1}{2})+\tan^{-1}(\dfrac{2}{5})$ is ?

$\tan^{-1} (5)$

$\tan^{-1}(\dfrac{1}{5})$

$\tan^{-1} (\dfrac{2}{3})$

$\tan^{-1} (\dfrac{8}{9})$

$\tan^{-1} (\dfrac{9}{8})$

Answer

$\tan^{-1} (\dfrac{9}{8})$

Explanation

$\tan^{-1}(\dfrac{1}{2})+\tan^{-1}(\dfrac{2}{5})$

[Here we know that $\tan^{-1}(x)+\tan^{-1}(y) =\tan^{-1}(\dfrac{x+y}{1-(x×y)})$

Now applying the above condition for the given question we can write,

$=\tan^{-1}(\dfrac{(1/2+2/5)}{1-((1/2)×(2/5))})$

$=\tan^{-1}(\dfrac{9}{8})$ (Ans..)

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