Question

if alpha beta are acute and roots of 2 tan squared theta + 35 tan theta + 2 is equals to zero then show that alpha + beta is equal to π/2​

Answer 1

Answer:

Step-by-step explanation:

1) We have,

$\alpha$ and $\beta$ are roots of equation $2tan^2(\theta)+tan(\theta)+2=0$

$tan(\alpha)$ and $tan(\beta)$ are roots of $2x^2+35x+2=0.$

2) We know,

Sum of roots=-35/2

$=>tan(\alpha)+tan(\beta)=\frac{-35}{2}$

Product of roots =2/2=1

$tan(\alpha)*tan(\beta)=1$

3) That is,

$tan(\alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)*tan(\beta)}\\ \\=\frac{\frac{-35}{2}}{1-1}=-\infty\\ \\ =>\alpha+\beta=\frac{\pi }{2}$

Hence, Required value is  $\pi/2$.

Answer 2

$2 { \tan }^{2} theta + 35 \tan \: theta \: + 2 = 0 \\ \tan( \alpha ) + \tan( \beta ) = \frac{ - 35}{2} \\ \tan( \alpha ) \tan( \beta ) = \frac{2}{2} = 1 \\ \tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) } \\ = \frac{ \frac{ - 35}{2} }{1 - 1} = \frac{ \frac{ - 35}{2} }{0} = - \infty \\ \tan( \alpha + \beta ) = - \infty \\( \alpha + \beta) = \frac{\pi}{2}$