If $y(\alpha ) = \sqrt {2\left( {\frac{{\tan \alpha + \cot \alpha }}{{1 + {{\tan }^2}\alpha }}} \right) + \frac{1}{{{{\sin }^2}\alpha }}} ,\alpha \in \left( {\frac{{3{\rm{\pi }}}}{4},{\rm{\pi }}} \right),$ then ${\rm{ }}\frac{{dy}}{{d\alpha }}$ at $\alpha = \frac{{5{\rm{\pi }}}}{6}$ is

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

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Analyzing +4 | -1

y(\alpha ) = \sqrt {2\left( {\frac{{\tan \alpha + \cot \alpha }}{{1 + {{\tan }^2}\alpha }}} \right) + \frac{1}{{{{\sin }^2}\alpha }}} ,\alpha \in \left( {\frac{{3\mathrm{\pi }}}{4},\mathrm{\pi }} \right),

then

\mathrm{ }\frac{{dy}}{{d\alpha }}

\alpha = \frac{{5\mathrm{\pi }}}{6}

View answer and explanation

Correct option: D

Option D (4) is correct because after simplifying the expression for y(

\alpha

) and differentiating, we find that

\frac{{dy}}{{d\alpha}}

\alpha = \frac{{5\pi}}{{6}}

is equal to 4.

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