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If t a ntheta=-1/(sqrt(5)) and theta lie...

If `t a ntheta=-1/(sqrt(5))` and `theta` lies in the IVB quadrant, then the value of `costheta` is a.`(sqrt(5))/(sqrt(6))` b. `2/(sqrt(6))` c. `1/2` d. `1/(sqrt(6))`

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Given `theta` lies in first quadrant `=> cos theta>0`
`tan theta =1/sqrt5`
As we know that
`sec^2 theta =1 + tan^2=1+(1/sqrt5)=1+(1/5)=6/5`
...
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