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If cottheta+tantheta=xand sectheta-costh...

If `cottheta+tantheta=xand sectheta-costheta=y`, then

A

`xsintheta.costheta=1`

B

`sin^2theta=ycostheta`

C

`(x^2y)^(1//3)+(xy^2)^(1//3)=1`

D

`(x^2y)^(2//3)-(xy^2)^(2//3)=1`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
`cottheta+tantheta=x`
`or costheta/sintheta+sintheta/costheta=x`
`or 1=xsinthetacostheta`
Now, `sectheta-costheta=y`
`1/costheta-costheta=y`
`or sin^2tehta=ycostheta`
`Now, x^2y=1/(sin^2thetacos^2theta).sin^2theta/costheta=1/cos^3theta`
`and xy^2=sin^3theta/cos^3theta`
`:. (x^2y)^(2//3)=1/cos^2theta-sin^2theta/cos^2theta=1`
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