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Mathematics Question on Trigonometric Functions

1sin⁡θ−3cos⁡θ=\frac{1}{\sin\theta}- \frac{\sqrt{3}}{\cos \theta}=

A

4cos⁡(π3−θ)sin⁡2θ=\frac{4 \cos\left(\frac{\pi}{3} -\theta\right) }{\sin2\theta} =

B

4sin⁡(π3−θ)sin⁡2θ=\frac{4 \sin\left(\frac{\pi}{3} -\theta\right) }{\sin2\theta} =

C

4cos⁡(π3+θ)sin⁡2θ=\frac{4 \cos\left(\frac{\pi}{3} + \theta\right) }{\sin2\theta} =

D

4sin⁡(π3+θ)sin⁡2θ=\frac{4 \sin\left(\frac{\pi}{3} + \theta\right) }{\sin2\theta} =

Answer

4cos⁡(π3+θ)sin⁡2θ=\frac{4 \cos\left(\frac{\pi}{3} + \theta\right) }{\sin2\theta} =

Explanation

Solution

1sin⁡θ−3cos⁡θ=cos⁡θ−sin⁡θ3sin⁡θcos⁡θ\frac{1}{\sin \theta }- \frac{\sqrt{3}}{ \cos \theta} = \frac{ \cos\theta -\sin \theta\sqrt{3} }{\sin \theta \cos \theta}
Putting 1=rcos⁡ϕ1 =r \cos \phi and 3=rsin⁡ϕ,\sqrt{3} =r \sin\phi, we get
∴  r=1+3=2\therefore \, \, r =\sqrt{1+3} =2 and tan⁡ϕ=31=tan⁡π3\tan\phi = \frac{\sqrt{3}}{1} =\tan \frac{\pi}{3}
⇒ ϕ=π3\Rightarrow \, \phi =\frac{\pi}{3}
∴  1sin⁡θ−3cos⁡θ=rcos⁡ϕcos⁡θ−rsin⁡ϕsin⁡θsin⁡θcos⁡θ\therefore \, \, \frac{1}{\sin \theta } - \frac{\sqrt{3} }{\cos \theta }= \frac{r \cos \phi \cos\theta - r \sin\phi \sin \theta }{\sin\theta \cos \theta}
=2r(cos⁡ϕcos⁡θ−sin⁡ϕsin⁡θ)2sin⁡θcos⁡θ= \frac{2r\left(\cos \phi \cos \theta - \sin \phi \sin \theta \right)}{2\sin \theta \cos \theta }
=2.2cos⁡(ϕ+θ)sin⁡2θ=4cos⁡(π3+θ)sin⁡2θ= \frac{2.2 \cos \left(\phi + \theta\right) }{\sin 2 \theta } = \frac{4 \cos \left(\frac{\pi}{3} + \theta\right) }{\sin2\theta }