If `sin(x+20^0)=2sinxcos40^0,w h r ex in (0,pi/2),` then which of the following hold(s) good? `cos2x=1/2` (b) `cos e c4x=2` `secx/2=sqrt(6)-sqrt(2)` (d) `tanx/2=(2-sqrt(3))`

Options

A. `cos 2x=1//2`
B. `co sec 4x=2`
C. `sec((x)/(2))=sqrt(6)-sqrt(2)`
D. `tan((x)/(2))=(2-sqrt(3))`

Answer

A, C, D

Explanation

`sin x cos 20^(@)+cos x sin 20^(@)=2sin x cos 40^(@)`
or `sin20^(@)cosx=sinx(2cos40^(@)-cos20^(@))`
or `tanx=(sin20^(@))/(2cos40^(@)-cos20^(@))`
`=(sin20^(@))/(cos40^(@)+cos40^(@)-cos20^(@))`
`=(sin 20^(@))/(cos40^(@)+2sin30^(@)sin(-10^(@)))`
`=(sin20^(@))/(sin50^(@)-sin10^(@))=(sin20^(@))/(2cos30^(@)sin 20^(@))`
`=(1)/sqrt(3) "or" x-30^(@)`