In a triangle ABC, `C = 90^@`, then the equation whose roots are `tanA, tanB `is

Explanation

In right angled `DeltaABC, angleC=90^(@)`
`therefore" "tan(A+B)=(tanA+tanB)/(1-tanAtanB)`
`rArr" "(1)/(0)=(tanA+tanB)/(1-tanAtanB)`
`rArr" "tanAtanB=1`
`" "tanA+tanB=(sinA)/(cosA)+(sinB)/(cosB)" "…(i)`
`" "=(sinA)/(cosA)+(sin(90^(@)-A))/(cos(90-A))" "[because angleC=90^(@), angleB=90^(@)-A]`
`" "=(sinA)/(cosA)+(cosA)/(sinA)`
`" "=(sin^(2)A+cos^(2)A)/(sinA*cosA)`
`" "=(1)/(sinA*cosA)=(2)/(2*sinA*cosA)`

`" "=(2)/(sin2A)" "[thereforesin2x=2sinxcosx]`
So, the required equation is `x^(2)-((2)/(sinA))x+1`.