Verified Solution • 2025–26 Session moderate 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 logcosxcotx+4logsinxtanx=1,x∈(0,π2)\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right)logcosxcotx+4logsinxtanx=1,x∈(0,2π) , is sin−1(α+β2)\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)sin−1(2α+β) , where α\alphaα , β\betaβ are integers, then α+β\alpha+\betaα+β is equal to : Multiple Choice Options A 3 B 6 C 4 D 5 Step-by-Step Verified Solution logcosxcotx+4logsinxtanx=1{\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1logcosxcotx+4logsinxtanx=1 ⇒logcosxcotx−4logsinxcotx=1\Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1⇒logcosxcotx−4logsinxcotx=1 ⇒1−logcosxsinx−4−4logsinxcosx=1\Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1⇒1−logcosxsinx−4−4logsinxcosx=1 Let logcosxsinx=t{\log _{\cos x}}\sin x = tlogcosxsinx=t t+4t=4t + {4 \over t} = 4t+t4=4 ⇒t=2\Rightarrow t = 2⇒t=2 sinx=cos2x\sin x = {\cos ^2}xsinx=cos2x ⇒sinx=1−sin2x\Rightarrow \sin x = 1 - {\sin ^2}x⇒sinx=1−sin2x ⇒sin2x+sinx−1=0\Rightarrow {\sin ^2}x + \sin {x^{ - 1}} = 0⇒sin2x+sinx−1=0 ⇒sinx=−1 ± 52\Rightarrow \sin x = {{ - 1\, \pm \,\sqrt 5 } \over 2}⇒sinx=2−1±5 as x∈(0,π2)x \in \left( {0,{\pi \over 2}} \right)x∈(0,2π) sinx=5−12\sin x = {{\sqrt 5 - 1} \over 2}sinx=25−1 x=sin−1(−1+52)x = {\sin ^{ - 1}}\left( {{{ - 1 + \sqrt 5 } \over 2}} \right)x=sin−1(2−1+5) ⇒α=−1,β=5\Rightarrow \alpha = - 1,\beta = 5⇒α=−1,β=5 α+β=4\alpha + \beta = 4α+β=4 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring