The value of sin^-1displaystyle12/13 + cos^-14/5 + tan^-163/16 is:

The correct answer is Option (2) → \(\pi\)

Let \(\alpha = \sin^{-1}\frac{12}{13}\) and \(\beta = \cos^{-1}\frac{4}{5}\). Both are acute angles. From the right triangles with sides 5, 12, 13 and 3, 4, 5:

    \( \displaystyle \tan\alpha = \frac{12}{5}, \quad \tan\beta = \frac{3}{4} \)

By the tangent addition formula:

    \( \displaystyle \tan(\alpha + \beta) = \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5}\times\frac{3}{4}} = \frac{\frac{63}{20}}{-\frac{4}{5}} \)

    \( \displaystyle = -\frac{63}{16} \)

Since \(\alpha + \beta\) lies between 0 and \(\pi\) and its tangent is negative, \(\alpha + \beta\) is obtuse:

    \( \displaystyle \alpha + \beta = \pi - \tan^{-1}\frac{63}{16} \)

Hence

    \( \displaystyle \sin^{-1}\frac{12}{13} + \cos^{-1}\frac{4}{5} + \tan^{-1}\frac{63}{16} = \pi \)

Note: This topic was not in the CUET 2026 syllabus.