Q.39. If \( \sec \theta=4 x \) and \( \tan \theta=\frac{4}{x},(x \neq 0) \) then the value of \( 8\left(x^{2}-\frac{1}{x^{2}}\right) \) is : SSC CHSL \( 05 / 07 / 2019 \) (Afternoon)

To solve the problem, we start with the definitions of secant and tangent. Given that \( \sec \theta = 4x \) and \( \tan \theta = \frac{4}{x} \), we can use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \). 1. We know: \[ \sec^2 \theta = (4x)^2 = 16x^2 \] \[ \tan^2 \theta = \left(\frac{4}{x}\right)^2 = \frac{16}{x^2} \] 2. Using the identity: \[ 16x^2 = 1 + \frac{16}{x^2} \] 3. Multiply both sides by \( x^2 \) to eliminate the fraction: \[ 16x^4 = x^2 + 16 \] Rearranging gives: \[ 16x^4 - x^2 - 16 = 0 \] Letting \( y = x^2 \), we can rewrite this as: \[ 16y^2 - y - 16 = 0 \] 4. Next, we solve for \( y \) using the quadratic formula: \[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 16 \cdot (-16)}}{2 \cdot 16} = \frac{1 \pm \sqrt{1 + 1024}}{32} = \frac{1 \pm 32}{32} \] This gives us: \[ y_1 = 1 \quad \text{and} \quad y_2 = -\frac{31}{32} \quad (\text{discarded, } y \geq 0) \] 5. Therefore, \( x^2 = 1 \). Now, we need to find \( 8(x^2 - \frac{1}{x^2}) \): \[ 8\left(x^2 - \frac{1}{x^2}\right) = 8\left(1 - \frac{1}{1}\right) = 8(0) = 0 \] So the final answer is: \[ \boxed{0} \]