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QJEEM23Mabf86: JEE Main 2023 Mathematics - Class 12 JEEM23Mabf86 Solution: If the solution of the equation , is , where , are integers,…

If the solution of the equation log⁡cos⁡xcot⁡x+4log⁡sin⁡xtan⁡x=1,x∈(0,π2)\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right) , is sin⁡−1(α+β2)\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right) , where α\alpha , β\beta are integers, then α+β\alpha+\beta is equal to :

Multiple Choice Options

  • A 3
  • B 6
  • C 4
  • D 5

Step-by-Step Verified Solution

log⁡cos⁡xcot⁡x+4log⁡sin⁡xtan⁡x=1{\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1

⇒log⁡cos⁡xcot⁡x−4log⁡sin⁡xcot⁡x=1\Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1

⇒1−log⁡cos⁡xsin⁡x−4−4log⁡sin⁡xcos⁡x=1\Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1

Let log⁡cos⁡xsin⁡x=t{\log _{\cos x}}\sin x = t

t+4t=4t + {4 \over t} = 4

⇒t=2\Rightarrow t = 2

sin⁡x=cos⁡2x\sin x = {\cos ^2}x

⇒sin⁡x=1−sin⁡2x\Rightarrow \sin x = 1 - {\sin ^2}x

⇒sin⁡2x+sin⁡x−1=0\Rightarrow {\sin ^2}x + \sin {x^{ - 1}} = 0

⇒sin⁡x=−1 ± 52\Rightarrow \sin x = {{ - 1\, \pm \,\sqrt 5 } \over 2}

as x∈(0,π2)x \in \left( {0,{\pi \over 2}} \right)

sin⁡x=5−12\sin x = {{\sqrt 5 - 1} \over 2}

x=sin⁡−1(−1+52)x = {\sin ^{ - 1}}\left( {{{ - 1 + \sqrt 5 } \over 2}} \right)

⇒α=−1,β=5\Rightarrow \alpha = - 1,\beta = 5

α+β=4\alpha + \beta = 4

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