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Let -1/6 < theta < -pi/12 Suppose ...

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

A

`2(sec theta-tan theta)`

B

`2 sec theta`

C

`-2 tan theta`

D

`3 tan theta`

Text Solution

Verified by Experts

The correct Answer is:
C
Here, `x^(2)2x sec theta+1=0 "has roots"alpha_(1)and beta_(1).`
`therefore alpha_(1), beta_(1)=(2sectheta+-sqrt(4sec^(2)theta-4))/(2xx1)=(2sec theta+-2tan theta)/(2)`
Since, `thetain(-(pi)/(6),(pi)/(12)),`
`i.e. theta in IV "quadrant"=(2 sec theta+-2 tan theta)/(2)`
`therefore alpha_(1)=sec theta-tantheta and beta_(1)=sectheta+tan theta[as alpha_(1)gtbeta_(1)]`
and `x^(2)+2x tan theta-1=0 "has roots alpha_(1)and beta_(2).`
`i.e. alpha_(2).beta_(2)(-2tan+-sqrt(4tan^(2)theta+4))/(2)`
`thereforealpha_(2)=-tantheta+sec theta`
`and beta_(2)=-tantheta -sec theta" "[as alpha_(2)gtbeta_(2)]`
Thus, `alpha_(1)+beta_(2)=-2tantheta`
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