Let `-1/6 < theta < -pi/12` Suppose `alpha_1 and beta_1`, are the roots of the equation `x^2-2xsectheta + 1=0` and `alpha_2 and beta_2` are the roots of the equation `x^2 + 2xtantheta-1=0`. If `alpha_1 > beta_1` and `alpha_2 >beta_2`, then `alpha_1 + beta_2` equals

Options

A. `2(sec theta-tan theta)`
B. `2 sec theta`
C. `-2 tan theta`
D. `0`

Answer

A

Explanation

`:'x^(2)-2x sec theta a+1=0impliesx -sec theta +- tan theta`
and `-(pi)/6 lt theta lt -(pi)/12`
`impliessec (-(pi)/6)gtsec theta gt sec (-(pi)/12)`
or `sec((pi)/6)gtsec theta gt sec ((pi)/12)`
and `tan(-(pi)/6)lt tan theta lt tan(-(pi)/12)`
`implies-tan ((pi)/6)lt tan theta lt -tan((pi)/12)`
or `tan ((pi)/6)gt-tan theta gt tan ((pi)/12)`
`:'alpha_(1),beta_(1)`
`:'alpha_(1),beta_(1)` are roots of `x^(2)-2x sec theta=1=0` and `alpha_(1)gtbeta_(1)`
`:. alpha_(1)=sec theta-tan theta` and `beta_(1)=sec theta+tan theta`
`impliesalpha_(2),beta_(2)` are roots of `x^(2)+2xtan theta-1=0`
and `alpha_(2)gtbeta_(2)`
`:.alpha_(2)=-tan theta+sec theta`
and `beta_(2)=-tan theta-sec theta`
Hence `alpha_(1)+beta_(2)=-2tan theta`