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If sec `alpha and alpha` are the roots of `x^2-p x+q=0,` then (a) `p^2=q(q-2)` (b) `p^2=q(q+2)` (c)`p^2q^2=2q` (d) none of these

Text Solution

Verified by Experts

The correct Answer is:
`npi +(pi)/(6), n in Z`
We have ,
`sqrt(3)+ i =(a+ ib) (c+id)`
`rArr ac -bd = sqrt(3) and ad + bc = 1`
Now, `tan^(-1)((b)/(a)) + tan^(-1)((d)/(c)) = tan^(-1)((b)/(a)+(d)/(c))/(1-(b)/(a)(d)/(c))`
`=tan ^(-1).(bc+ad)/(ac-db)`
`=tan^(-1)(1)/(sqrt(3)) = npi+(pi)/(6), n in Z`
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