Verified Solution • 2025–26 Session moderate

QJEEM20Ma1729: JEE Main 2020 Mathematics - Class 12 JEEM20Ma1729 Solution: If :

If y(α)=2(tan⁡α+cot⁡α1+tan⁡2α)+1sin⁡2α,α∈(3π4,π)y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)

dydα  at α=5π6is{{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is :

Multiple Choice Options

  • A 4
  • B -4
  • C 43{4 \over 3}
  • D - 14{1 \over 4}

Step-by-Step Verified Solution

y(α)=2(tan⁡α+cot⁡α1+tan⁡2α)+1sin⁡2αy\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}

= 2(1+tan⁡2αtan⁡α(1+tan⁡2α))+1sin⁡2α\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}}

= 2(cos⁡αsin⁡α)+1sin⁡2α\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}

= sin⁡2α+1sin⁡2α\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}}

= ∣sin⁡α+cos⁡α∣∣sin⁡α∣{{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}

At α\alpha = 5π6{{5\pi } \over 6}

∣sin⁡α+cos⁡α∣{\left| {\sin \alpha + \cos \alpha } \right|} = -(sin α\alpha + cos α\alpha )

and |sin α\alpha | = sin α\alpha

∴\therefore y( α\alpha ) = −(sin⁡α+cos⁡α)sin⁡α{{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}

= -1 - cot α\alpha

∴\therefore dydα{{dy} \over {d\alpha }} = cosec2 α\alpha

So dydα{{{dy} \over {d\alpha }}} at α\alpha = 5π6{{5\pi } \over 6} ,

= cosec2 5π6{{5\pi } \over 6} = 4

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