IIM-K (BMSAT) 2025Geometry > Easy(1+cosθ)sinθ\frac{(1+ \cos \theta)}{\sin \theta}sinθ(1+cosθ)(1−cosθ)cosθ\frac{(1-\cos \theta)}{\cos \theta}cosθ(1−cosθ)(1−cosθ)sinθ\frac{(1-\cos \theta)}{\sin \theta}sinθ(1−cosθ)(1−sinθ)cosθ\frac{(1-\sin \theta)}{\cos \theta}cosθ(1−sinθ)✅ 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 isGiven below are two statements: Statement I: cot30∘+1cot30∘−1=2(cos30∘+1)\frac{\cot 30^{\circ}+1}{\cot 30^{\circ}-1}=2\left(\cos 30^{\circ}+1\right)cot30∘−1cot30∘+1=2(cos30∘+1) Statement II : 2sin45∘cos45∘−tan45∘cot45∘=02 \sin 45^{\circ} \cos 45^{\circ}-\tan 45^{\circ} \cot 45^{\circ}=02sin45∘cos45∘−tan45∘cot45∘=0If cosαcos \alphacosα + cosβcos \betacosβ = 1 then the maximum value of sinα−sinβsin \alpha - sin \betasinα−sinβ is