Discussion Forum



1)

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 $ and $Q-P\neq \phi$

B) $Q\neq\subset P$

C) $P\neq\subset Q$

D) P=Q

Answer:

Option D

Explanation:

$P=( \theta: \sin \theta-\cos \theta= \sqrt{2} \cos \theta)$

$\Rightarrow$  $\cos \theta ( \sqrt{2}+1)= \sin \theta$

$\Rightarrow$  $\tan \theta = \sqrt{2}+1$ ...(i)

$Q=( \theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta)$

$\Rightarrow$   $\sin \theta(\sqrt{2}-1)=\cos \theta$

$\Rightarrow$  $\tan \theta= \frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

   $= (\sqrt{2}+1)$....(ii)

$\therefore$ P=Q