Statement (S1): $\sin 55^{\circ}+\sin 53^{\circ}-\sin 19^{\circ}-\sin 17^{\circ}=$ $\cos 2^{\circ}$.
Statement (S2): Range of $\frac{1}{3-\cos 2 x}$ is $\left[\frac{1}{4}, \frac{1}{2}\right]$
Which one of the following is correct?
Both (S1) and (S2) are true
Both (S1) and (S2) are false
$(\mathrm{S} 1)$ is true, (S2) is false
$(\mathrm{S} 1)$ is false, (S2) is true
Solution
I. Since, $\sin 55^{\circ}+\sin 53^{\circ}-\sin 19^{\circ}-\sin 17^{\circ}$
$\begin{aligned} & =\left(\sin 55^{\circ}-\sin 17^{\circ}\right)+\left(\sin 53^{\circ}-\sin 19^{\circ}\right) \\ & =2 \cos 36^{\circ} \cdot \sin 19^{\circ}+2 \cos 36^{\circ} \cdot \sin 17^{\circ} \\ & =2 \cos 36^{\circ}\left(2 \sin 18^{\circ} \cos 1^{\circ}\right)=2\left(\frac{\sqrt{5}-1}{4}\right) \cdot 2\left(\frac{\sqrt{5}+1}{4}\right) \cdot \cos 1^{\circ} \\ & =\cos 1^{\circ} .\end{aligned}$
II. Let $f(x)=\frac{1}{3-\cos 2 x}$
$\begin{aligned}
& \text { Since, }-1 \leq-\cos 2 x \leq 1 \\
& \Rightarrow-1+3 \leq 3-\cos 2 x \leq 1+3 \\
& \Rightarrow \frac{1}{4} \leq f(x) \leq \frac{1}{2}
\end{aligned}$
So, I is false and II is true.