First slide

Trigonometric Identities

Question

$1 + \sin x + \sin^{2} x + \dots to \infty = 4 + 2 \sqrt{3}$ If

Moderate

A

$x = \frac{2 π}{3} or, \frac{π}{3}$
B

$x = \frac{7 π}{6}$
C

$x = \frac{π}{6}$
D

$x = \frac{π}{4}$

Solution

We have,
$1 + \sin x + \sin^{2} x + \dots + to \infty = (\sqrt{3} + 1)^{2}$
$\begin{array}{l} \Rightarrow \frac{1}{1 - \sin x} = (\sqrt{3} + 1)^{2} \\ \Rightarrow 1 - \sin x = \frac{1}{(\sqrt{3} + 1)^{2}} \\ \Rightarrow 1 - \sin x = \frac{(\sqrt{3} - 1)^{2}}{4} \\ \Rightarrow \sin x = 1 - (\frac{4 - 2 \sqrt{3}}{4}) \\ \Rightarrow \sin x = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{π}{3} or, \frac{2 π}{3} \end{array}$

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