First slide

Compound angles

Question

let $α & β$ be two real roots of the equation $(k + 1) \tan^{2} x - \sqrt{2} . λ \tan x = (1 - k)$ , where $k (\neq - 1)$ and $λ$ are real numbers. If $\tan^{2} (α + β) = 50$ then the value of $λ$ is

Easy

A

$10 \sqrt{2}$
B

$5 \sqrt{2}$
C

10
D

5

Solution

Given that $\tan α a n d \tan β a r e r o o t s o f t h e e q u a t i o n$
$(k + 1) \tan^{2} x - \sqrt{2} λ \tan x + (k - 1) = 0$
$\Rightarrow$ sum of roots = $\tan α + \tan β = \frac{\sqrt{2} λ}{k + 1}$
and, product of roots = $\tan α \tan β = \frac{k - 1}{k + 1}$
Now $\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} = \frac{\frac{\sqrt{2} λ}{k + 1}}{1 - \frac{k - 1}{k + 1}} = \frac{\sqrt{2} λ}{2} = \frac{λ}{\sqrt{2}}$
$\Rightarrow \tan^{2} (α + β) = \frac{λ^{2}}{2} = 50$
$\Rightarrow λ = 10$

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