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Let (3pi)/4 < theta < pi and sqrt(2 cot ...

Let `(3pi)/4 < theta < pi` and `sqrt(2 cot theta+1/sin^2 theta) = k - cot theta` then `k=`

A

1

B

`-1`

C

0

D

`1/2`

Text Solution

Verified by Experts

The correct Answer is:
B
We have,
`sqrt(2 cottheta+(1)/(sin^(2)theta))=k cottheta`
`implies sqrt(2 cottheta+cosec^(2)theta)=k-cottheta`
`impliessqrt(2cot theta+1cot^(2)theta)=k-cottheta`
`implies[1+cottheta]=k-cottheta`
`implies-1(1+cot theta)=k-cottheta[(3pi)/(4)lt thetaltpiimpliescot thetalt-1implies-oolt1+cotthetalt0]`
`-1-cottheta=k -cottheta`
`impliesk=-1`
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