Let [ ] and [ ] be two real roots of the equation (k + 1)tan2x - [ 2 ] .

Question

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:

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

Solutions · 1