Question

If $\tan^{2} \theta, \sec^{2} \theta$ are the roots of $ax^{2} + bx + c = 0$ then $b^{2} - a^{2} =$

A $4ac$ B $a^{2}$ C $4bc$ D $4ab$

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Answer

To solve for the value of $ b^2 - a^2 $ given that $\tan^2 \theta$ and $\sec^2 \theta$ are the roots of the quadratic equation $ ax^2 + bx + c = 0 $, follow these steps:

  1. Sum and Product of Roots:

    • Recognize that for a quadratic equation $ ax^2 + bx + c = 0 $ with roots $\alpha$ and $\beta$, the sum of the roots is given by: $$ \alpha + \beta = -\frac{b}{a} $$ and the product of the roots is: $$ \alpha \beta = \frac{c}{a} $$
    • Here, let $\alpha = \sec^2 \theta$ and $\beta = \tan^2 \theta$.
  2. Key Identity:

    • Use the trigonometric identity: $$ \sec^2 \theta - \tan^2 \theta = 1 $$
  3. Equation Setup:

    • Using the identities for roots, we substitute: $$ \alpha + \beta = \sec^2 \theta + \tan^2 \theta = -\frac{b}{a} $$ $$ \alpha \beta = \sec^2 \theta \cdot \tan^2 \theta = \frac{c}{a} $$
  4. Relation Application:

    • Using the relation $(\alpha + \beta)^2 = (\alpha - \beta)^2 + 4\alpha \beta$, substitute the known values: $$ \left( -\frac{b}{a} \right)^2 = (\sec^2 \theta - \tan^2 \theta)^2 + 4 \frac{c}{a} $$
    • As $\sec^2 \theta - \tan^2 \theta = 1$: $$ \left( -\frac{b}{a} \right)^2 = 1^2 + 4 \frac{c}{a} $$
  5. Simplify:

    • Simplify the above equation: $$ \frac{b^2}{a^2} = 1 + \frac{4c}{a} $$
    • Cross-multiply to clear the fraction: $$ b^2 = a^2 + 4ac $$
  6. Final Result:

    • Rearrange to find $b^2 - a^2$: $$ b^2 - a^2 = 4ac $$

Thus, the value of $ b^2 - a^2 $ is 4ac.

Answer: A


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