Let `P={theta:sintheta-costheta=sqrt2costheta}and Q={theta: sintheta+costheta=sqrt2sintheta}` be two sets. Then

Options

A. `Q notinP`
B. `P notinQ`
C. `P sub Q and Q-P ne phi`
D. `P=Q`

Answer

D

Explanation

Let Q `in` P. Then,
`thetainPhArrsin theta-costheta=sqrt2costheta`
`hArr1-2sinthetacostheta=2cos^(2)theta`
`hArr2 sin thetacos theta=2 cos^(2)theta`
hArr`1+2sin thetacos theta=2-2 cos^(2)theta`
`hArr(sin theta+cos theta)^(2)=2 sin ^(2)theta`
`hArr sintheta+cos theta=sqrt2sintheta`
`hArrtheta in Q.`
hence, `P =Q.`