Question

Let P $ = \left\{ {0:\sin \theta - \cos \theta = \sqrt 2 \cos \theta } \right\}$ and Q $ = \left\{ {0:\sin \theta + \cos \theta = \sqrt 2 \sin \theta } \right\}$be two sets. Then:
$
  a)\,P \subset Q\,\,{\text{and}}\,Q - P \ne \emptyset \\
  b)\,Q \not\subset P \\
  c)\,P = Q \\
  d)\,P \not\subset Q \\
 $

Answer 1

Hint: Given P$ = \left\{ {0:\sin \theta - \cos \theta = \sqrt 2 \cos \theta } \right\}$ divide it by $\cos \theta $, you will get the value of , similarly do it with set Q, you will get your answer.

Complete step-by-step answer:
Hence question is given two sets P and Q. P is defined as $ = \left\{ {0:\sin \theta - \cos \theta = \sqrt 2 \cos \theta } \right\}$and
Q is defined as$ = \left\{ {0:\sin \theta + \cos \theta = \sqrt 2 \sin \theta } \right\}$
Now let us solve it separately.
Now if we look in the set P
P\[ = \left\{ {0:\sin \theta - \cos \theta = \sqrt 2 \cos \theta } \right\}\]
It is the relation between $\sin \theta \,\,\& \,\,\cos \theta $which indicates we can get $\tan \theta $easily.
So now let's solve. Here,
\[
  \sin \theta - \cos \theta = \sqrt 2 \cos \theta \\
  \sin \theta = \sqrt 2 \cos \theta + \cos \theta \\
  \sin \theta = \sqrt 2 + 1\left( {\cos \theta } \right) \\
  \tan \theta = \sqrt 2 + 1 \\
 \]
So here we can also define set P as
P $ = \left\{ {0:\tan \theta = \sqrt 2 + 1} \right\}$
Now, for set Q again we are provided with the same function of $\sin \theta \,\,\& \,\,\cos \theta $which indicates we can get $\tan \theta $easily. Given
Q$ = \left\{ {0:\sin \theta + \cos \theta = \sqrt 2 \sin \theta } \right\}$
So now lets solve the equation
$
  \cos \theta = \sqrt 2 \sin \theta - \sin \theta \\
  \cos \theta = \sqrt 2 - 1\left( {\sin \theta } \right) \\
  \tan \theta = \dfrac{1}{{\sqrt 2 - 1}} \\
 $
A denominator has a root term, so we can rationalize upon multiplying and dividing its conjugate.
$
  \tan \theta = \dfrac{1}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}} \\
  \tan \theta = \dfrac{{\sqrt 2 + 1}}{{{{\sqrt 2 }^2} - 1}} \\
  \tan \theta = \dfrac{{\sqrt 2 + 1}}{{2 - 1}} \\
  \tan \theta = \sqrt 2 + 1 \\
 $
So we get set Q $ = \left\{ {0:\tan \theta = \sqrt 2 + 1} \right\}$
Hence we got that set P $ = \left\{ {0:\tan \theta = \sqrt 2 + 1} \right\}$
So both sets are equal, which means P=Q is our right choice.

So option C is correct.

Note: If we are given P$ \subseteq $Q then this means P is a subset of Q or Q is contained in P. this means every element of set P is present in Q. If P $ \not\subset $Q then none of the elements of set P belong to set Q.