"Let $\alpha$ and $\beta$ be two real roots of the equation (k + 1)tan2x - $\sqrt 2$ . $\lambda$tanx = (1 - k), where k($\ne$ - 1) and $\lambda$ are real numbers. if tan2 ($\alpha$ + $\beta$) = 50, then a value of $\lambda$ is:"

Question

Accepted Answer

"Let tan$\alpha$ and tan$\beta$ are the roots of

(k + 1)tan²x - $\sqrt 2$ . $\lambda$tanx - (1 - k) = 0

$\therefore$ tan$\alpha$ + tan$\beta$ = ${{\sqrt 2 \lambda } \over {k + 1}}$

and an$\alpha$.tan$\beta$ = ${{k - 1} \over {k + 1}}$

Now tan($\alpha$ + $\beta$)

= ${{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}$

= ${{{{\lambda \sqrt 2 } \over {k + 1}}} \over {1 - {{k - 1} \over {k + 1}}}}$

= ${{{\lambda \sqrt 2 } \over 2}}$ = ${{\lambda \over {\sqrt 2 }}}$

Given ${{{{\lambda ^2}} \over 2}}$ = 50

$\Rightarrow$ $\lambda$ = 10"

Let $\alpha$ and $\beta$ be two real roots of the equation
(k + 1)tan²x - $\sqrt 2$ . $\lambda$tanx = (1 - k), where k($\ne$ - 1) and $\lambda$ are real numbers. if tan² ($\alpha$ + $\beta$) = 50, then a value of $\lambda$ is:

Solution

About this question

Drill 25 more like these. Every day. Free.

Let $\alpha$ and $\beta$ be two real roots of the equation (k + 1)tan2x - $\sqrt 2$ . $\lambda$tanx = (1 - k), where k($\ne$ - 1) and $\lambda$ are real numbers. if tan2 ($\alpha$ + $\beta$) = 50, then a value of $\lambda$ is:

Solution

About this question

More Complex Numbers and Quadratic Equations questions

Drill 25 more like these. Every day. Free.

Let $\alpha$ and $\beta$ be two real roots of the equation
(k + 1)tan²x - $\sqrt 2$ . $\lambda$tanx = (1 - k), where k($\ne$ - 1) and $\lambda$ are real numbers. if tan² ($\alpha$ + $\beta$) = 50, then a value of $\lambda$ is: