If 7 cos2θ + 5 sin2θ - 6 = 0, (0° < θ < 90°), then what is the value of ?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. + 1.
7cos2θ + 5sin2θ - 6 = 0 (0° < θ < 90°)
⇒ 7(1 - sin2θ) + 5sin2θ - 6 = 0
⇒ 7 - 7sin2θ + 5sin2θ - 6 = 0
⇒ sin2θ = 1/2
⇒ sinθ = 1/√2
⇒ θ = 45°
Now,
= √[(√2 + 1)/(√2 - 1)]
= √[(√2 + 1)2]
= √2 + 1