If $ \tan \frac{\alpha}{2} $ and $ \tan \frac{\beta}{2} $ are the roots of $ 8x^2 - 26x + 15 = 0 $, then $ \cos(\alpha + \beta) $ is

Question

If $ \tan^{-1}\left(\frac{1}{1+1\cdot2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}(x) $, then $ x $ is equal to:

Answer 1

Understanding the 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: \[ \cos(\alpha+\beta) = -\frac{627}{725} \] Final Answer:
Answer = $-\frac{627}{725}$

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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The Correct Option is B

Solution and Explanation

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