Practicing Success
If 2 sin2θ + 3 cosθ = 3, 0o < θ < 90o, then the value of (sec2θ +cot2 θ) is |
$3\frac{2}{3}$ $3\frac{1}{3}$ $4\frac{1}{3}$ $4\frac{1}{2}$ |
$4\frac{1}{3}$ |
We are given that :- 2 sin²θ + 3 cosθ = 3 { we know, sin²θ + cos²θ = 1 } 2 ( 1 - cos²θ ) + 3 cosθ = 3 2 - 2cos²θ + 3 cosθ = 3 2cos²θ - 3 cosθ + 1 = 0 2cos²θ - 2 cosθ - cosθ + 1 = 0 2 cosθ ( cosθ - 1 ) - 1 ( cosθ - 1 ) = 0 ( 2cosθ - 1 ) . ( cosθ - 1 ) = 0 ( cosθ - 1 ) = 0 is not possible because 0º < θ < 90º So, 2cosθ - 1 = 0 cosθ = \(\frac{1}{2}\) { we know, cos60º = \(\frac{1}{2}\) } So, θ = 60º Now, ( sec ²θ + cot ²θ ) = ( sec ²60º + cot ²60º ) = 4 + \(\frac{1}{3}\) = 4\(\frac{1}{3}\)
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