IPMAT Indore 2020Geometry > Hard11132\frac{3}{2}2322294\frac{9}{4}49✅ Correct Option: 3Related questions:For 0<θ<π40\lt\theta\lt\frac{\pi}{4}0<θ<4π, let a=((sinθ)sinθ)(log2cosθ),b=((cosθ)sinθ)(log2sinθ),c=((sinθ)cosθ)(log2cosθ)a=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \cos \theta\right), b=\left((\cos \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right), c=\left((\sin \theta)^{\cos \theta}\right)\left(\log _{2} \cos \theta\right)a=((sinθ)sinθ)(log2cosθ),b=((cosθ)sinθ)(log2sinθ),c=((sinθ)cosθ)(log2cosθ) and d=((sinθ)sinθ)(log2sinθ)d=\left((\sin \theta)^{\sin \theta}\right)\left(\log _{2} \sin \theta\right)d=((sinθ)sinθ)(log2sinθ). Then, the median value in the sequence a,b,c,da, b, c, da,b,c,d isIf sinα+sinβ=23\sin \alpha+\sin \beta=\frac{\sqrt{2}}{\sqrt{3}}sinα+sinβ=32 and cosα+cosβ=13\cos \alpha+\cos \beta=\frac{1}{\sqrt{3}}cosα+cosβ=31, then the value of (20cos(α−β2))2\left(20 \cos \left(\frac{\alpha-\beta}{2}\right)\right)^{2}(20cos(2α−β))2 is _________.If cosαcos \alphacosα + cosβcos \betacosβ = 1 then the maximum value of sinα−sinβsin \alpha - sin \betasinα−sinβ is