Q87 / 180 · KCET 2016 · Mathematics

If \( 1+\sin \theta+\sin ^{2} \theta+\ldots . \) upto \( \infty=2 \sqrt{3}+4 \), then \( \theta= \)

  • A
    \( \frac{\pi}{6} \)
  • B
    \( \frac{\pi}{4} \)
  • Correct answer
    C
    \( \frac{\pi}{3} \)
  • D
    \( \frac{3 \pi}{4} \)

Solution

Given that, $1+\sin \theta+\sin ^{2} \theta+\ldots+\infty=2 \sqrt{3}+4$ We know that $S_{\infty}=\frac{a}{1-r}$ Here $a=1$ and $r=\sin \theta$ $$ \begin{array}{l} \text { So, } \frac{1}{1-\sin \theta}=2 \sqrt{3}+4 \\ \Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \\ \Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \times \frac{2 \sqrt{3}-4}{2 \sqrt{3}-4} \\ \Rightarrow 1-\sin \theta=\frac{2 \sqrt{3}-4}{12-16}=\frac{2 \sqrt{3}-4}{-4} \\ \Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\ \Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\ \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3} \end{array} $$

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

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