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

The value of $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \ldots + \sin^2 85^{\circ} + \sin^2 90^{\circ} = $

Evaluate the sum: $\sum \frac{1}{1 + x^{a - b} + x^{a - c}}$

Let $x=a \sin ^\alpha \theta \cos ^{\alpha+1} \theta$ and $y=a \sin ^{\alpha+1} \theta \cos ^\alpha \theta$,where $\theta \neq \frac{n \pi}{2}$. If $\frac{(x^2+y^2)^m}{(xy)^n}$ is independent of $\theta$,then the relation between $\alpha, m$ and $n$ is:

If $\sec \theta \cosh y = \operatorname{cosec} x$ and $\operatorname{cosec} \theta \sinh y = \sec x$,then $\sinh ^2 y =$

The value of $\text{cosec}10^{\circ} - \sqrt{3} \text{sec}10^{\circ}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module