Geometry > TrigonometryEasyPrev37 of 50Next$\dfrac{\sin \theta}{(1+ \cos \theta)}$ is equal to(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:BMSAT Kozhikode 2025If 1+sin2θ=3sinθcosθ1 + \sin^2\theta = 3 \sin \theta \cos \theta1+sin2θ=3sinθcosθ, then tanθ\tan \thetatanθ can take valuesBMSAT Kozhikode 2025If a pole 6 m high casts a shadow 232\sqrt{3}23 m long on the ground, then the sun's elevation is2026If a pole is 6 meters high and casts a shadow 232\sqrt{3}23 meters long on the ground, then the sun's elevation is2026If cosα+cosβ+cosγ=3\cos\alpha + \cos\beta + \cos\gamma = 3cosα+cosβ+cosγ=3, then cos3α+cos3β+cos3γ\cos^3\alpha + \cos^3\beta + \cos^3\gammacos3α+cos3β+cos3γ will be2026In △ABC\triangle ABC△ABC, ∠A=90∘\angle A = 90^{\circ}∠A=90∘, if tanC=3\tan C = \sqrt{3}tanC=3, then find the value of sinB+cosC−cos2B\sin B + \cos C - \cos^2 BsinB+cosC−cos2B