Correct Answer - Option 4 : Neither 1 nor 2
Calculations :
Staement 1
sin θ = x + (1/x)
⇒ sin θ = (x2 + 1)/x
⇒ x.sin θ = x2 + 1
⇒ x2 - x.sin θ + 1 = 0
For any quadratic equation(ax2 + bx + c) real values exist when b2 > 4ac
So,
(- sin θ)2 > 4 × 1 × 1
⇒ sin2 θ > 4
⇒ sin θ > 2
We know that -1 ≤ sin θ ≤ 1
So no value exist for some real value of x.
Statement 2
cos θ = x + (1/x)
⇒ cos θ = (x2 + 1)/x
⇒ x.cos θ = x2 + 1
⇒ x2 - x.cos θ + 1 = 0
For any quadratic equation(ax2 + bx + c) real values exist when b2 > 4ac
So,
(- cos θ)2 > 4 × 1 × 1
⇒ cos2 θ > 4
⇒ cos θ > 2
We know that -1 ≤ cos θ ≤ 1
So no value exist for some real value of x.
So no statement is following.
∴ Option 4 is the correct choice.