यदि `6sin^(4)theta+3cos^(4)theta= 2` तो `[7"co s ec"^(6)theta+8sec^(6)theta]^(1//3)` का मान है:

Options

A. 2
B. 4
C. 8
D. 6

Answer

d

Explanation

`6sin^(4)theta+3cos^(4)theta=2`
`rArr 6sin theta^(4)+3(1-sin^(2)theta)^(2)=2`
`rArr6 sin^(4)theta+3+3sin^(4)theta-6sin^(2)theta=2`
`rArr9sin^(4)theta- 6sin^(2)theta+1=0`
`rArr(3sin^(2)theta-1)^(2)=0`
`rArr(3sin^(2)theta-1)^(2)=0`
`rArr3sin^(2)theta=1`
`rArr sintheta=(1)/sqrt(3)`

Now, `(7co s ec^(4)theta+8sec^(6)theta)^(1//3)`
`=(7xxsqrt(3))^(6)+8((sqrt3)/(sqrt(2)))^(6 )1//3`
`=(7xx27+8xx(27/(8)1//3`
`=(27xx8)1//3=6`