Let P = { theta : sin theta - cos theta = sqr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$. Rearranging gives $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$. Multiplying

Vedclass · Accepted Answer

$P = Q$

Vedclass · Answer

(C) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$. Rearranging gives $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$. Multiplying by $(\sqrt{2} - 1)$,we get $(\sqrt{2} - 1) \sin 	heta = (\sqrt{2} - 1)(\sqrt{2} + 1) \cos 	heta$,which simplifies to $(\sqrt{2} - 1) \sin 	heta = \cos 	heta$. Thus,$\sin 	heta = \cos 	heta + \sqrt{2} \sin 	heta - \sin 	heta$,which leads to $\sin 	heta + \cos 	heta = \sqrt{2} \sin 	heta$. This is the defining condition for set $Q$. Similarly,starting from $Q$,we can derive the condition for $P$. Therefore,$P = Q$.

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

Similar Questions

If $\frac{1}{6} \sin \theta, \cos \theta$ and $\tan \theta$ are in geometric progression,then the solution set of $\theta$ is

$16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ} = $

If $x \neq -y$ and $\sin x + \sin y = 3(\cos y - \cos x)$,then $\tan(x - y) =$

If $\cos x - \sin x = \sqrt{a} \sin x$,then $a \sin x + \cos x - \sin x = $

$\cos 12^{\circ} \cdot \cos 24^{\circ} \cdot \cos 36^{\circ} \cdot \cos 48^{\circ} \cdot \cos 72^{\circ} \cdot \cos 84^{\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module