Solving for sin(α + β) sec α sec β using Quadratic Roots

The problem asks us to find the value of the expression $\sin (\alpha + \beta) \sec \alpha \sec \beta$, given that $\tan \alpha$ and $\tan \beta$ are the roots of the quadratic equation $x^2 + bx + c = 0$, where $b \ne 0$.

Understanding the Relationship between Roots and Coefficients

For a general quadratic equation of the form $Ax^2 + Bx + C = 0$, the sum of the roots is given by $-B/A$ and the product of the roots is given by $C/A$. This is known as Vieta's formulas.

In our given equation, $x^2 + bx + c = 0$, we have $A=1$, $B=b$, and $C=c$. The roots are given as $\tan \alpha$ and $\tan \beta$.

Using Vieta's formulas:

• Sum of roots: $\tan \alpha + \tan \beta = \frac{-b}{1} = -b$
• Product of roots: $\tan \alpha \tan \beta = \frac{c}{1} = c$

We now have the sum and product of $\tan \alpha$ and $\tan \beta$ in terms of $b$ and $c$.

Simplifying the Expression sin(α + β) sec α sec β

Let's simplify the expression we need to evaluate: $\sin (\alpha + \beta) \sec \alpha \sec \beta$.

We can use standard trigonometric identities:

• The sum formula for sine: $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
• The reciprocal identity for secant: $\sec \theta = \frac{1}{\cos \theta}$

Substitute these into the expression:

$\sin (\alpha + \beta) \sec \alpha \sec \beta = (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \cdot \frac{1}{\cos \alpha} \cdot \frac{1}{\cos \beta}$

$= \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}$

Now, we can split the fraction into two terms:

$= \frac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta} + \frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}$

Cancel out the common terms in each fraction:

$= \frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, this simplifies to:

$= \tan \alpha + \tan \beta$

Finding the Value of the Expression

We have shown that $\sin (\alpha + \beta) \sec \alpha \sec \beta$ is equal to $\tan \alpha + \tan \beta$.

From Vieta's formulas applied to the given quadratic equation, we found that $\tan \alpha + \tan \beta = -b$.

Therefore, $\sin (\alpha + \beta) \sec \alpha \sec \beta = -b$.

Conclusion

The value of $\sin (\alpha + \beta) \sec \alpha \sec \beta$ is $-b$. This matches option 2.

Revision Table: Key Concepts Used

Additional Information: Quadratic Roots and Trigonometry

This problem is a good example of how concepts from different areas of mathematics, specifically algebra (quadratic equations and roots) and trigonometry, can be combined. When encountering problems involving trigonometric functions of angles that are roots of a polynomial, Vieta's formulas are often very useful.

The condition $b \ne 0$ is important because if $b=0$, the equation becomes $x^2 + c = 0$, and the sum of roots $\tan \alpha + \tan \beta$ would be 0. This would imply $-b=0$, which aligns, but the problem states $b \ne 0$ to ensure a non-zero sum of roots.

Also, for $\tan \alpha$ and $\tan \beta$ to be real roots, the discriminant of the quadratic equation ($b^2 - 4c$) must be non-negative ($b^2 - 4c \ge 0$).