The quadratic equation whose roots are sec⁡2θ and csc⁡θ can bea) x2−2x+2=0b) x2−5x+5=0c) x2−7x+7=0d) x2−9x+9=0

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Kendall Clark · Accepted Answer

Step 1As others have pointed out, you’ve assume that if u≥v then u2≥v2This is true if u,v are both non-negative, but if both are non-positive, then the inequality reverses, u2≤v2, and there is nothing we can say if u is positive and v negative.Another way to write the equation is as((cos2θ)x−1)((sin2θ)x−1)=(cos2θsin2θ)x2−x+1So the monic polynomial with these two roots hasp=q=1sin2θcos2θ=4sin22θThis is really the same as your first approach, but switches to sin and cos immediately.

Kendall Wilkinson · Accepted Answer

Step 1Recall that1sec2θ+1csc2θ=1. Ifx2+px+q=0thenx=x1=−p+p2−4q2orx=x2=−p−p2−4q2.So you want 1x1+1x2=1 You have1x1+1x2=2−p+p2−4q+2−p−p2−4q=−pqSo,1=−pqp=−q−p±p2+4p2={−1±3if p=2,⋯if p=5,⋯if p=7,⋯if p=9.When p=2, one of the solutions is negative and so cannot be sec⁡2θ or csc⁡2θ if θ is real. The other one is negative and so cannot be sec⁡2θ or csc⁡2θ if θ is real.Go on from there is a similar way.

The quadratic equation whose roots are \(\displaystyle{\sec{{2}}}\theta\)

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