Q.If \[ \tan \theta \cdot \cos 60^\circ = \frac{\sqrt{3}}{2} \] then find the value of \[ \sin(\theta - 15^\circ) \] given that \(0^\circ < \theta < 90^\circ\).

\[ \tan \theta \cdot \cos 60^\circ = \frac{\sqrt{3}}{2} \Rightarrow \tan \theta \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \Rightarrow \tan \theta = \frac{\sqrt{3}}{2} \cdot 2 = \sqrt{3} \Rightarrow \tan \theta = \tan 60^\circ \Rightarrow \theta = 60^\circ \] Therefore, \[ \sin(\theta - 15^\circ) = \sin(60^\circ - 15^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}} \]
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