Question:
If \( \tan \frac{\alpha}{2} \) and \( \tan \frac{\beta}{2} \) are the roots of \( 8x^2 - 26x + 15 = 0 \), then \( \cos(\alpha + \beta) \) is
If \( \tan \frac{\alpha}{2} \) and \( \tan \frac{\beta}{2} \) are the roots of \( 8x^2 - 26x + 15 = 0 \), then \( \cos(\alpha + \beta) \) is
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Use $t = \tan(\theta/2)$ to simplify trig expressions.
Updated On: Apr 23, 2026
- $\frac{627}{725}$
- $-\frac{627}{725}$
- $-1$
- None of these
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The Correct Option is B
Solution and Explanation
Concept:
Using $t = \tan(\theta/2)$ identity
Step 1: Let roots be $t_1, t_2$.
Step 2: Sum and product: \[ t_1 + t_2 = \frac{26}{8}, \quad t_1 t_2 = \frac{15}{8} \]
Step 3: Use identity: \[ \cos(\alpha+\beta) = \frac{1 - t_1^2 - t_2^2 + t_1^2 t_2^2}{(1+t_1^2)(1+t_2^2)} \]
Step 4: Express in terms of sum and product and simplify.
Step 5: Final value: \[ \cos(\alpha+\beta) = -\frac{627}{725} \] Conclusion:
Answer = $-\frac{627}{725}$
Step 1: Let roots be $t_1, t_2$.
Step 2: Sum and product: \[ t_1 + t_2 = \frac{26}{8}, \quad t_1 t_2 = \frac{15}{8} \]
Step 3: Use identity: \[ \cos(\alpha+\beta) = \frac{1 - t_1^2 - t_2^2 + t_1^2 t_2^2}{(1+t_1^2)(1+t_2^2)} \]
Step 4: Express in terms of sum and product and simplify.
Step 5: Final value: \[ \cos(\alpha+\beta) = -\frac{627}{725} \] Conclusion:
Answer = $-\frac{627}{725}$
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