`underset0overset(pi/3)int tantheta/(sqrt(2ksectheta))d theta=1-1/sqrt2,(kgt0)`, then the value of k is

Answer

2

Explanation

`int_(0)^(pi//3)(tantheta)/(sqrt(2k sec theta))d theta=1-(1)/(sqrt2),(k gt0)`
`int_(0)^(pi//3)(tan theta)/(sqrt(2k cos theta))d theta=(1)/(sqrt(2k))int_(0)^(pi//3)(sin theta)/(sqrt(cos theta))d theta`
`"Put "sqrt(costheta)=1`
`-(1)/(2sqrt(cos theta))sin theta d theta=dt, -(1)/(sqrt(2 cos theta))sin theta d theta=dt sqrt2 =-(1)/(sqrtk)int_(1)^((1)/(sqrt2))sqrt2dt=-(sqrt2)/(sqrtk)(-(1)/(sqrt2)-1)=1-(1)/(sqrt2)`
`rArr" "k=2`