$P=\left\{\theta:\sin \theta-\cos \theta = \sqrt{2}\cos \theta\right\}$ Let $P=\left\{\theta:\sin \theta-\cos \theta = \sqrt{2}\cos \theta\right\}$ and $Q=\left\{\theta:\sin \theta+\cos \theta=\sqrt{2}\sin \theta\right\}$ be two sets , Then
which one is Right.
$(a)\;\;\; P\subset Q$ and $Q-P\neq \phi\;\;\;\;\;\; (b)\;\;\; Q\not\subset P\;\;\;\;\;\; (c)\;\;\;P\not\subset Q\;\;\;\;\;\; (d)\;\;\; P=Q$
$\bf{My\; Try::}$ Given $\left(\sin \theta - \cos \theta\right) = \sqrt{2}\cos \theta\Rightarrow \left(\sin \theta - \cos \theta\right)^2 = 2\cos^2 \theta$
$\displaystyle = \sin^2 \theta +\cos^2 \theta -2\sin \theta \cos \theta = 2\cos^2 \theta \Rightarrow \sin^2 \theta -\cos^2 \theta -2\sin \theta \cos \theta = 0$
$\Rightarrow \cos^2 \theta+2\sin \theta \cdot \cos \theta = \sin^2 \theta$
Now Add $\sin^2 \theta$ on both side, we get
$\Rightarrow \cos^2 \theta+2\sin \theta \cdot \cos \theta +\sin^2 \theta = \sin^2 \theta+\sin^2 \theta$ 
So $\left(\sin \theta +\cos \theta\right) = \left(\sqrt{2}\cdot \sin \theta\right)^2$
So $\left(\sin \theta +\cos \theta\right) = \sqrt{2}\sin \theta$ and $\left(\sin \theta +\cos \theta\right) = -\sqrt{2}\sin \theta$
So $Q = \left\{\theta:\sin \theta +\cos \theta = \pm \sqrt{2}\sin \theta\right\}$
My doubt is How can we say $P = Q$ because here set $Q$ contain $\pm $ sign.
and answer is also given $Q$, please explain me,
Thanks
 A: Unfortunately, you cannot get rid of the $\pm$ sign.  When you squared the set expression for $P$, the resulting set was not equivalent to $P$; in general, you cannot assume $\{x \mid f(x) = g(x)\} = \{x \mid f(x)^2 = g(x)^2\}$.  Usually the second set will be larger.
Here is an approach that avoids squaring by dividing instead:
$$\begin{align}
P &= \{\theta \mid \sin \theta - \cos \theta = \sqrt 2 \cos \theta\}\\
  &= \{\theta \mid \tan \theta - 1 = \sqrt 2 \} \quad\quad\text{ see note 1}\\
  &= \{\theta \mid \tan \theta = 1 + \sqrt 2 \} \\
\end{align}$$
$$\begin{align}
Q &= \{\theta \mid \sin \theta + \cos \theta = \sqrt 2 \sin \theta\}\\
&= \{\theta \mid 1 + \cot \theta = \sqrt 2\}\\
&= \{\theta \mid \tan \theta = \frac{1}{\sqrt 2 - 1}\}\\
&= \{\theta \mid \tan \theta = \frac{1}{\sqrt 2 - 1}\frac{\sqrt 2 + 1}{\sqrt 2 + 1}\}\\
&= \{\theta \mid \tan \theta = \sqrt 2 + 1\}\\
\end{align}$$
Nicely, they are the same set.  Note 1 : dividing by $\cos \theta$ can potentially lose any solutions of the form $\{\theta \mid \cos \theta = 0\}$ from $P$, so we have to double check those.  Applying $\cos \theta = 0$ to $P$, you get $P|_{\cos \theta = 0} = \{\theta \mid \sin \theta = 0\}$, and since $\sin \theta = 0$ and $\cos \theta = 0$ don't occur at the same time, the division doesn't lose any elements of the set $P$.
