Question

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 ,

• A. P $\subset$ $Q-P$ = $\varphi$
• B. Q $\subset$ P
• C. P $\subset$ Q
• D. $P=Q$

Answer 1

Answer: (D)

For set $P$,
$\Rightarrow \sin \theta -\cos \theta =\sqrt{2}\cos \theta$
$\Rightarrow \tan \theta -1=\sqrt{2}$
[dividing both sides by $\cos \theta$ ]
$\Rightarrow \tan \theta =\sqrt{2}+1...$(i)
For set $Q$
$\sin \theta +\cos \theta =\sqrt{2}\sin \theta$
$\Rightarrow 1+\cot \theta =\sqrt{2}$
[dividing both sides by $\sin \theta$ ]
$\Rightarrow \cot \theta =\sqrt{2}-1$
$\Rightarrow \tan \theta =\sqrt{2}+1...(ii)$
Hence, $P=Q$
[from Eqs. (i) and (ii)]