Question

\( \alpha, \beta \) are the roots of the equation \( \sin^2 x + b\sin x + c = 0 \). If \( \alpha + \beta = \frac{\pi}{2} \) then \( b^2 - 1 = \)

• A. c
• B. 2c
• C. \( c^2 \)
• D. \( 4c^2 \)

Hint:

Always clarify whether "roots" refers to the variable \( x \) or the variable of the quadratic substitution (e.g., \( \sin x \)). Here, \( \alpha, \beta \) are values of \( x \), so \( \sin \alpha, \sin \beta \) are the roots of the algebraic equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

The given equation is quadratic in terms of \( \sin x \). This means the roots of the quadratic equation \( t^2 + bt + c = 0 \) are the values \( \sin \alpha \) and \( \sin \beta \).
Step 2: Key Formula or Approach:

For a quadratic \( At^2 + Bt + C = 0 \) with roots \( x_1, x_2 \): \[ x_1 + x_2 = -B/A, \quad x_1 x_2 = C/A \] Identity: \( (\sin \theta + \cos \theta)^2 = 1 + 2\sin\theta\cos\theta \).
Step 3: Detailed Explanation:

The quadratic equation is \( (\sin x)^2 + b(\sin x) + c = 0 \). Since \( \alpha \) and \( \beta \) are the roots (values of \( x \)), the actual roots of the quadratic in \( t = \sin x \) are \( t_1 = \sin \alpha \) and \( t_2 = \sin \beta \). From the properties of quadratic equations: Sum of roots: \( \sin \alpha + \sin \beta = -b \) Product of roots: \( \sin \alpha \sin \beta = c \) Given \( \alpha + \beta = \frac{\pi}{2} \), we have \( \beta = \frac{\pi}{2} - \alpha \). Substitute this into the expression for the second root: \[ t_2 = \sin \beta = \sin\left(\frac{\pi}{2} - \alpha\right) = \cos \alpha \] So the roots are \( \sin \alpha \) and \( \cos \alpha \). Substituting back into the sum and product relations: 1. \( \sin \alpha + \cos \alpha = -b \) 2. \( \sin \alpha \cos \alpha = c \) We need to find \( b^2 - 1 \). Square the first relation: \[ (\sin \alpha + \cos \alpha)^2 = (-b)^2 \] \[ \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = b^2 \] Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) and substituting the product relation: \[ 1 + 2(c) = b^2 \] Rearranging terms: \[ b^2 - 1 = 2c \]
Step 4: Final Answer:

The value of \( b^2 - 1 \) is \( 2c \).