MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Use right-triangle representation of inverse trigonometric functions to compare both sides of the equation.Explanation:Let $\theta = \cos^{-1} x$, so $\cos \theta = x$.Then $\sin \theta = \sqrt{1 - x^2}$.Therefore, $\cot(\cos^{-1} x) = \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{\sqrt{1-x^2}}$.Now let $\phi = \tan^{-1}\left(\frac{a}{\sqrt{b^{2}-a^{2}}}\right)$, so $\tan \phi = \frac{a}{\sqrt{b^{2}-a^{2}}}$.For this angle, perpendicular $= a$ and base $= \sqrt{b^{2}-a^{2}}$.Hypotenuse $= \sqrt{a^{2} + (b^{2} - a^{2})} = b$.Hence, $\sec \phi = \frac{\text{hypotenuse}}{\text{base}} = \frac{b}{\sqrt{b^{2}-a^{2}}}$.Given $\cot(\cos^{-1} x) = \sec(\tan^{-1}\frac{a}{\sqrt{b^{2}-a^{2}}})$,So, $\frac{x}{\sqrt{1-x^{2}}} = \frac{b}{\sqrt{b^{2}-a^{2}}}$.Squaring both sides: $\frac{x^{2}}{1-x^{2}} = \frac{b^{2}}{b^{2}-a^{2}}$.Cross-multiplying: $x^{2}(b^{2} - a^{2}) = b^{2}(1 - x^{2})$.This simplifies to $x^{2}(2b^{2} - a^{2}) = b^{2}$.Taking the positive root, $x = \frac{b}{\sqrt{2b^{2} - a^{2}}}$.Answer:$x = \frac{b}{\sqrt{2b^{2} - a^{2}}}$Hence, the correct option is Option D.