JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 17)
Let $$P = \{ \theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta \} $$ and $$Q = \{ \theta :\sin \theta + \cos \theta = \sqrt 2 \sin \theta \} $$ be two sets. Then
$$P \subset Q$$ and $$Q - P \ne \emptyset $$
$$Q \not\subset P$$
$$P \not\subset Q$$
$$P = Q$$
Explanation
$$P = \{ \theta :\sin \theta - \cos \theta = \sqrt 2 \cos \theta \} $$
$$ \Rightarrow \cos \theta \left( {\sqrt 2 + 1} \right) = \sin \theta $$
$$ \Rightarrow \tan \theta = \sqrt 2 + 1$$ ..... (i)
$$Q = \{ \theta :\sin \theta + \cos \theta = \sqrt 2 \sin \theta \} $$
$$ \Rightarrow \sin \theta \left( {\sqrt 2 - 1} \right) = \cos \theta $$
$$ \Rightarrow \tan \theta = {1 \over {\sqrt 2 - 1}} \times {{\sqrt 2 + 1} \over {\sqrt 2 + 1}}$$
$$ = \left( {\sqrt 2 + 1} \right)$$ ...... (ii)
$$\therefore$$ $$P = Q$$
Comments (0)
