Question

The general solution of the equation $\frac{1-\sin x+\ldots+(-1)^{n} \sin ^{n} x+\ldots \ldots}{1+\sin x+\ldots+\sin ^{n} x+\ldots}=\frac{1-\cos 2 x}{1+\cos 2 x}$

• A. $(-1)^{n}(\pi / 3)+n \pi$
• B. $(-1)^{n}(\pi / 6)+n \pi$
• C. $(-1)^{n+1}(\pi / 6)+n \pi$
• D. $(-1)^{n-1}(\pi / 3)+n \pi$

Answer 1

Answer: (B)

$\frac{1-\sin x+\ldots+(-1)^{n} \sin ^{n} x+\ldots \ldots}{1+\sin x+\ldots+\sin ^{n} x+\ldots}=\frac{1-\cos 2 x}{1+\cos 2 x}$
$\Rightarrow \frac{1}{1+\sin x} \times \frac{1-\sin x}{1}=\frac{2 \sin ^{2} x}{2 \cos ^{2} x}$
$\Rightarrow \cos ^{2} x-\cos ^{2} x \sin x=\sin ^{2} x+\sin ^{3} x\,\,(\sin x+1 \neq 0)$
$\Rightarrow 2 \sin ^{2} x+\sin x-1=0$
$\Rightarrow \sin x=-1$ or $\sin x=1 / 2$
Since $\sin x \neq-1$
we have $\sin x=1 / 2=\sin (\pi / 6)$
$\Rightarrow x=n \pi+(-1)^{n}(\pi / 6), n \in Z$