The value of $\mathop {\lim }\limits_{\alpha \to 0} \frac{{\left[ {{\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\sec \alpha } \right) + {{\cot }^{ - 1}}\left( {\tan \alpha } \right) + {{\cot }^{ - 1}}\cos \left( {{{\sin }^{ - 1}}\alpha } \right)} \right]}}{\alpha }$ is

JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

The value of

\lim\limits_{\alpha \to 0} \frac{{\left[ {\mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\sec \alpha } \right) + {{\cot }^{ - 1}}\left( {\tan \alpha } \right) + {{\cot }^{ - 1}}\cos \left( {{{\sin }^{ - 1}}\alpha } \right)} \right]}}{\alpha }

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Correct option: C

- 2

Option C (

-2

) is correct because after simplifying the trigonometric inverse functions and applying L'Hopital's rule, the limit evaluates to -2.