JEE PYQ: Trigonometric Equations And Inequations Question 8
Question 8 - 2020 (07 Jan Shift 1)
Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1) \tan^2 x - \sqrt{2} \cdot \lambda \tan x = (1-k)$, where $k(\ne -1)$ and $\lambda$ are real numbers. If $\tan^2(\alpha + \beta) = 50$, then a value of $\lambda$ is:
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(1) $10\sqrt{2}$
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(2) $10$
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(3) $5$
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(4) $5\sqrt{2}$
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Answer:
Solution
$\tan(\alpha+\beta) = \frac{\lambda}{\sqrt{2}}$
$\tan^2(\alpha+\beta) = \frac{\lambda^2}{2} = 50 \Rightarrow \lambda = 10$
BETA
