sin θ=tanθ ⟹ sin θ1= sin θcos θ\sin\;\theta = \tan\theta \implies\; \frac{\sin\;\theta}{1} = \; \frac{\sin\;\theta}{\cos\;\theta}sinθ=tanθ⟹1sinθ=cosθsinθ
Equating, we have
cosθ=1 ⟹ θ=cos−11\cos\theta = 1 \implies \theta = \cos^{-1} 1cosθ=1⟹θ=cos−11
= 0∘0^{\circ}0∘
