KeerthanaPosted on If tanθ⋅cos60∘=32 an heta cdot cos 60^{circ}=frac{sqrt{3}}{2}tanθ⋅cos60∘=23, then the value of sin(θ−sin ( heta-sin(θ− 15∘15^{circ}15∘ ) is a1 b12frac{1}{sqrt{2}}21 c12frac{1}{2}21 d32frac{sqrt{3}}{2}23 Answer : Option BExplanation : tanθ⋅cos60∘=32 an heta cdot cos 60^{circ}=frac{sqrt{3}}{2}tanθ⋅cos60∘=23 ⇒tanθ×12=32Rightarrow an heta imes frac{1}{2}=frac{sqrt{3}}{2}⇒tanθ×21=23 ⇒tanθ=3=tan60∘Rightarrow an heta=sqrt{3}= an 60^{circ}⇒tanθ=3=tan60∘ ⇒θ=60∘Rightarrow heta=60^{circ}⇒θ=60∘ ∴sin(θ−15∘)=sin(60∘−15∘)=sin45∘=12 herefore sin left( heta-15^{circ} ight)=sin left(60^{circ}-15^{circ} ight)=sin 45^{circ}=frac{1}{sqrt{2}}∴sin(θ−15∘)=sin(60∘−15∘)=sin45∘=21 Rate This:NaN / 5 - 1 votesAdd comment
