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Q.

If tanθ  and secθ  are the roots of the equation ax2+bx+c=0 then:

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a

a4=b2(−2ac+b2)

b

b4=a2(2ac+a2)

c

a4=b2(−4ac+b2)

d

b4=a2(4ac+a2)

answer is C.

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Detailed Solution

tanθ+secθ=−b/a,  tanθsecθ=c/a ∴      secθ−tanθ=−a/b             [∵  sec2θ−tan2θ=1]⇒    secθ=−(a2+b2)2ab,   tanθ=(a2−b2)2ab⇒    (a2+b2)(b2−a2)4a2b2=ca⇒     b4−a4=4acb2
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