Let `0^(@)ltthetalt45^(@)`. Which one of the following is correct ?

Options

A. `sin^(2)theta+cos^(6)theta=sin^(6)theta+cos^(2)theta`
B. `"cosec"^(2)theta+cot^(6)theta="cosec"^(6)thetacot^(2)theta`
C. `sin^(2)theta-cos^(4)theta=sin^(4)theta+cos^(2)theta`
D. `"cosec"^(2)theta+cot^(4)theta="cosec"^(4)theta+cot^(2)theta`

Answer

D

Explanation

In such a problem , we have to check option , one- by - one
`sin^(2)theta+cos^(6)theta=sin^(6)theta+cos^(2)theta`
`rArrsin^(6)theta-cos^(6)theta=(sin^(2)theta)^(2)-(cos^(2)theta)^(2)`
Taking L . H . S,
`=(sin^(2)theta-cos^(2)theta)(sin^(4)theta+sin^(2)thetacos^(2)theta+cos^(4)theta)`
`(because a^(3)-b^(3)=(a-b)(a^(2)+ab+b^(2)))`
`=(sin^(2)theta-cos^(2)theta)(sin^(4)theta+cos^(4)theta+sin^(2)thetacos^(2)theta)`
`=(sin^(2)theta-cos^(2)theta)(sin^(2)theta+cos^(2)theta)-2sin^(2)thetacos^(2)theta+sin^(2)thetacos^(2)theta`
`=sin^(2)theta-cos^(2)theta-sin^(2)thetacos^(2)theta`
Which is not equal to R .H . S ., `sin^(2)theta-cos^(2)theta`
Option (a ) is not correct .
( b ) `therefore"cosec"^(6)theta-cot^(6)theta`
`("cosec"^(2)theta-cot^(2)theta)[("cosec"^(2)theta-cot^(2)theta)^(2)+("cosec "thetacottheta)]`
`therefore` Option (b ) is also not correct.
(c ) `sin^(4)theta+cos^(4)theta=(sin^(2)theta+cos^(2)theta)^(2)-2sin^(2)thetacos^(2)theta`
`=1-2sin^(2)thetacos^(2)theta`.
Which is not equal to `sin^(2)theta-cos^(2)theta`.
Hence , option ( c) is also not correct .
(d) `("cosec"^(2)theta+cot^(4)theta)`
`="cosec"^(2)theta+("cosec"^(2)theta-1)^(2)`
`="cosec"^(2)theta+"cosec"^(4)theta+1-2" cosec"^(2)theta`
`="cosec"^(4)theta+1-"cosec"^(2)theta`
`="cosec"^(4)theta+cot^(2)theta`
Thus option (d ) is correct .