Let P={theta:sintheta-costheta=sqrt2cost...

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

`psubQ and Q-P=phi`

`QcancelsubP`

`PcancelsubQ`

P=Q

Verified by Experts

The correct Answer is:

In set P, `sintheta=(sqrt2+1)costheta`
`or tantheta=sqrt2+1`
In set Q, `(sqrt2-1)sintheta=costheta`
`or tan thetaa=1/(sqrt2-1)=sqrt2+1`
`rArr P=Q`

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Let P={theta:sintheta-costheta=sqrt2cos theta}and Q={theta:sintheta+costheta=sqrt2sintheta} be two ses. Then,

`Q notinP`

`P notinQ`

`P sub Q and Q-P ne phi`

`P=Q`

`sqrt2cos theta`

`sqrt2 sintheta`

`sqrt2(cos theta+sintheta)`

none of these

`sqrt(2)+1`

`sqrt(2)-1`

`sqrt(3)-1`

`sqrt(3)+1`