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Trigonometrical Equations
hard
माना $P =\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ तथा $Q =\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$ दो समुच्चय हैं, तो
A
$P \subset Q$ and $Q - P \ne \phi $
B
$Q \not\subset P$
C
$P = Q$
D
$P \not\subset Q$
(JEE MAIN-2016)
Solution
$\sin \,\theta \, – \,\cos \,\theta \,\, = \,\,\sqrt 2 \,\cos \,\theta $
$ \Rightarrow \,\sin \,\theta \, = \,\cos \,\theta \,\, + \,\,\sqrt 2 \,\cos \,\theta $
$ = \,(\sqrt 2 \, + \,1)\cos \,\theta \, = \,\left( {\frac{{2\, – \,1}}{{\sqrt 2 \, – \,1}}} \right)\,\cos \,\theta $
$ \Rightarrow \,(\sqrt 2 \, – \,1)\sin \,\theta = \,\cos \,\theta $
$ \Rightarrow \,\sin \,\theta \, + \,\cos \,\theta \,\, = \,\,\sqrt 2 \,\cos \,\theta $
$\therefore \,\,P\, = \,Q$
Standard 11
Mathematics
