JIPMATGeometry > Easy11112\frac122134\frac344314\frac1441✅ Correct Option: 2Related questions:JIPMAT 2024Given 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∘=0JIPMAT 2023In △ABC,∠B=90∘,BC=5 cm,AC−AB=1 cm\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}, \mathrm{BC}=5 \ \mathrm{cm}, \mathrm{AC}-\mathrm{AB}=1 \mathrm{~cm}△ABC,∠B=90∘,BC=5 cm,AC−AB=1 cm, then 1+sin(C)1+cos(C)\frac{1+\sin (\mathrm{C})}{1+\cos (\mathrm{C})}1+cos(C)1+sin(C) isJIPMAT 2024If sinθ+cosθ=72\sin \theta+\cos \theta=\frac{\sqrt{7}}{2}sinθ+cosθ=27, then (sinθ−cosθ)(\sin \theta-\cos \theta)(sinθ−cosθ) is equal to :