Aspire Faculty ID #13513 · Topic: JAMIA MILLIA ISLAMIA MCA 2024 · Just now
JAMIA MILLIA ISLAMIA MCA 2024

$\tan^{-1}!\left(\dfrac{1}{2}\right) + \tan^{-1}!\left(\dfrac{1}{3}\right) =$

Solution

$\tan^{-1}a + \tan^{-1}b = \tan^{-1}!\left(\dfrac{a + b}{1 - ab}\right)$ Here, $a = \dfrac{1}{2}$, $b = \dfrac{1}{3}$ $\Rightarrow \dfrac{a + b}{1 - ab} = \dfrac{\frac{5}{6}}{1 - \frac{1}{6}} = 1$ $\Rightarrow \tan^{-1}(1) = \dfrac{\pi}{4}$

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