Question

If $\tan \theta .\cos {60}^o=\frac{\sqrt{3}}{2}$ then find the value of $\sin (\theta -{15}^o)$.

Answer 1

Given,

$\tan \theta .\cos {60}^o=\frac{\sqrt{3}}{2}$

or, $\tan \theta .\frac{1}{2}=\frac{\sqrt{3}}{2}$

or, $\tan \theta =\sqrt{3}$

or, $\theta ={60}^o$. [ Since $\tan {60}^o=\sqrt{3}$]

Now

$\sin (\theta -{15}^o)$

$=\sin ({60}^o-{15}^o)$

$=\sin {45}^o$

$=\frac{1}{\sqrt{2}}$