If `cosx=tany ,coty=tanz ,cotz=tanx ,` then the value of `sinx` is (a)`2cos18^0` (b) `cos18^0` (c) `sin18^0` (d) `2sin18^0`

Options

A. `2cos 18^(@)`
B. `cos18^(@)`
C. `sin18^(@)`
D. `2 sin18^(@)`

Answer

D

Explanation

We have `cos x=tan y`
or `cos^(2)x=tan^(2)y`
`=sec^(2)y-1`
`=cot^(2)z-1 [because cos y=tan z, secy=cotz]`
`rArr 1+cos^(2)x=cot^(2)z=(tan^(2)x)/(1-tan^(2)x)`
`=(sin^(2)x)/(cos^(2)x-sin^(2)x)`
`rArr 2sin^(4)x-6sin^(2)x+2=0`
or `sin^(2)x=(3-sqrt(5))/(2)=((sqrt(5)-1)/(2))^(2)`
or `sin x=(sqrt(5)-1)/(2)=2sin18^(@)`.