Let `-pi/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

C

Explanation

`x^(2) - 2x sec tan + 1 = 0" "....(i)`
`rArr" "x^(2) - x [(sec theta + tan theta)+(sec theta-tan theta)] + sec^(2) theta - tan^(2) theta = 0`
`rArr" "x^(2)-[(sec theta + tan theta)+(sec theta - tan theta)]+(sec theta + tan theta)(sec theta - tan theta) = 0`
`rArr" "x = sec theta + tan theta, x = sec theta - tan theta`
It is given that `-(pi)/(6) lt theta lt -(pi)/(12) and alpha_(1), beta_(1)` are roots of (i) such that `a_(1) gt beta_(1)`.
`therefore" "alpha_(1) = sec theta - tan theta and beta_(1) = sec theta + tan theta`
Now, `x^(2) + 2x tan theta - 1 = 0" "....(i)`
`rArr" "x^(2) - [(sec theta - tan theta)+(-tan theta - sec theta)]x+(sec theta-tan theta)(-tan theta-sec theta) = 0`
`rArr" "x = sec theta - tan theta, x = - tan theta - sec theta`
It is given that `theta in (-pi//6, -pi//12) and alpha_(2), beta_(2)` are roots of equation (ii) such that `alpha_(2) gt beta_(2)`.
`therefore" "alpha_(2) = sec theta - tan theta and beta_(2) = - tan theta - sec theta`
Hence `alpha_(1) + beta_(2) = sec theta - tan theta - tan theta - sec theta = -2 tan theta`.