Q63 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

The general solution of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

  • Correct answer
    A
    $2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}$
  • B
    $n \pi+(-1)^n \frac{\pi}{4}+\frac{\pi}{12}$
  • C
    $2 n \pi \pm \frac{\pi}{4}-\frac{\pi}{12}$
  • D
    $n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{12}$

Solution

Given, $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ Let $\quad \sqrt{3}-1=r \sin \alpha$ $\ldots(i)$ and $\quad \sqrt{3}+1=r \cos \alpha$ $\therefore r^2\left(\sin ^2 \alpha+\cos ^2 \alpha\right)=(\sqrt{3}-1)^2+(\sqrt{3}+1)^2$ $\Rightarrow \quad r^2=3+1-2 \sqrt{3}+3+1+2 \sqrt{3}$ $\Rightarrow r^2=8 \Rightarrow r=2 \sqrt{2}$ and $\tan \alpha=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{1-1 / \sqrt{3}}{1+1 / \sqrt{3}}$ $=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\tan \left(\frac{\pi}{12}\right) \Rightarrow \alpha=\frac{\pi}{12}$ From Eq. (i), we have $r \sin \alpha \sin \theta+r \cos \alpha \cos \theta=2$ $\Rightarrow \quad r \cos (\theta-\alpha)=2$ $\Rightarrow \quad 2 \sqrt{2} \cos (\theta-\alpha)=2$ $\Rightarrow \quad \cos (\theta-\alpha)=\frac{1}{\sqrt{2}}$ $\Rightarrow \quad \theta-\alpha=2 n \pi \pm \frac{\pi}{4}$ $\Rightarrow \quad \theta=2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}$

Hence, the correct answer is option: 'A'.

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