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

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.