Mht CetInverse Trigonometric FunctionsMHT CET 2026 11th April Evening ShiftMCQ+2 / -02026If $\cot(\cos^{-1} x) = \sec\left(\tan^{-1} \dfrac{a}{\sqrt{b^2 - a^2}}\right)$, then the value of $x$ isA$\dfrac{b}{\sqrt{2b^2 + a^2}}$B$\dfrac{\sqrt{2b^2 - a^2}}{b}$C$\dfrac{\sqrt{2b^2 + a^2}}{b}$D$\dfrac{b}{\sqrt{2b^2 - a^2}}$Check AnswerClear SelectionReveal AnswerShow Explanation