Trigonometric EquationHard
Question
If 4sin2θ + 4cosec2θ = 17, then the value of tan θ is -
Options
A.4 + √3
B.4 - √3
C.4 + √15
D.4 - √15
Solution
4sin 2θ + 
= 17
4t2 - 17t + 4 = 0
t =
&t = 4
sin 2θ =
& sin2 θ = 4 (rejected)
tan2θ - 8tanθ + 1 = 0
tanθ =
tan θ = 4 +
tan θ = 4 -
= 174t2 - 17t + 4 = 0
t =
&t = 4sin 2θ =
& sin2 θ = 4 (rejected)tan2θ - 8tanθ + 1 = 0
tanθ =
tan θ = 4 +
tan θ = 4 -
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