Let P={theta: sin theta-cos theta=sqrt{2} cos | vedclass

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

Question

$P=\{	heta: \sin 	heta-\cos 	heta=\sqrt{2} \cos 	heta\}$

$Q=\{	heta: \sin 	heta+\cos 	heta=\sqrt{2} \sin 	heta\}$

From P

$\sin 	heta

Vedclass · Accepted Answer

$P=Q$

Vedclass · Answer

$P=\{	heta: \sin 	heta-\cos 	heta=\sqrt{2} \cos 	heta\}$

$Q=\{	heta: \sin 	heta+\cos 	heta=\sqrt{2} \sin 	heta\}$

From P

$\sin 	heta-\cos 	heta=\sqrt{2} \cos 	heta$

$\sin 	heta=(\sqrt{2}+1) \cos 	heta$

$\frac{\sin 	heta}{\cos 	heta}=(\sqrt{2}+1)$

$	an 	heta=(\sqrt{2}+1)$

$	ext { from } Q$

$\sin 	heta+\cos 	heta=\sqrt{2} \sin 	heta$

$\sin 	heta(\sqrt{2}-1)=\cos 	heta$

$\frac{\sin 	heta}{\cos 	heta}=(\sqrt{2}-1)$

$	an 	heta=(\sqrt{2}-1)$

$	an 	heta=\frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$

$	herefore P=Q$

Hence, option $(D)$ is correct answer.

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 $A =$ [$x:x$ is a multiple of $3$] and $B =$ [$x:x$ is a multiple of $5$], then $A -B$ is ($\bar A$ means complement of $A$)

The shaded region in the given figure is

Consider the sets $A$ and $B$ of $A=\{2,4,6,8\}$ and $B=\{6,8,10,12\}$ Find $A \cap B .$

If $A=\{1,2,3,4\}, B=\{3,4,5,6\}, C=\{5,6,7,8\}$ and $D=\{7,8,9,10\} ;$ find

$B \cup C \cup D$

If $A = \{2, 3, 4, 8, 10\}, B = \{3, 4, 5, 10, 12\}, C = \{4, 5, 6, 12, 14\}$ then $(A \cap B) \cup (A \cap C)$ is equal to