Q.13. Let , and . Then is equal to:

Q: What is the correct answer for JEE Main Mathematics Q13?

The correct answer is B. Sol. Given $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$. Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy < 1$. Here $x = 1-\alpha$ and $y = 1-\beta$. $\tan^{-1}\left(\frac{(1-\alpha) + (1-\beta)}{1-(1-\alpha)(1-\beta)}\right) = \frac{\pi}{4}$. $\frac{2 - (\alpha+\beta)}{1 - (1 - \alpha - \beta + \alpha\beta)} = \tan(\frac{\pi}{4}) = 1

Q: Q.13. Let , and . Then is equal to:

The correct answer is B. Sol. Given $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$. Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy < 1$. Here $x = 1-\alpha$ and $y = 1-\beta$. $\tan^{-1}\left(\frac{(1-\alpha) + (1-\beta)}{1-(1-\alpha)(1-\beta)}\right) = \frac{\pi}{4}$. $\frac{2 - (\alpha+\beta)}{1 - (1 - \alpha - \beta + \alpha\beta)} = \tan(\frac{\pi}{4}) = 1

Question

Q.13. Let ,  and . Then  is equal to:

FirstInTest · Accepted Answer

Sol. Given $	an^{-1}(1 - \alpha) + 	an^{-1}(1 - \beta) = \frac{\pi}{4}$. Using the formula $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$ when $xy < 1$. Here $x = 1-\alpha$ and $y = 1-\beta$. $	an^{-1}\left(\frac{(1-\alpha) + (1-\beta)}{1-(1-\alpha)(1-\beta)}ight) = \frac{\pi}{4}$. $\frac{2 - (\alpha+\beta)}{1 - (1 - \alpha - \beta + \alpha\beta)} = 	an(\frac{\pi}{4}) = 1$. $\frac{2 - (\alpha+\beta)}{\alpha + \beta - \alpha\beta} = 1$. $2 - (\alpha+\beta) = \alpha + \beta - \alpha\beta$. $2 = 2(\alpha+\beta) - \alpha\beta$. We are given $\beta = \frac{1}{3\alpha}$, so $\alpha\beta = \frac{1}{3}$. Substitute $\alpha\beta = \frac{1}{3}$ into the equation: $2 = 2(\alpha+\beta) - \frac{1}{3}$. $2 + \frac{1}{3} = 2(\alpha+\beta)$. $\frac{7}{3} = 2(\alpha+\beta)$. $\alpha+\beta = \frac{7}{6}$. We need to find $6(\alpha+\beta)$. $6(\alpha+\beta) = 6\left(\frac{7}{6}ight) = 7$. We should check the condition $xy < 1$. $(1-\alpha)(1-\beta) < 1$. $1 - (\alpha+\beta) + \alpha\beta < 1$. $1 - \frac{7}{6} + \frac{1}{3} < 1$. $1 - \frac{7}{6} + \frac{2}{6} < 1$. $1 - \frac{5}{6} < 1$. $\frac{1}{6} < 1$. This condition is satisfied.

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6

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7

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8

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9

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