Question

tan α. tan β = a and α + β = π/6, then tan α and tan β are roots of the quadratic equation :
a) √3 x² - (1 - a)x + √3 a = 0
b) √3 x² - (1 + a)x + √3 a = 0
c) √3 x² + (1 - a)x - √3 a = 0
d) √3 x² + (1 - a)x + √3 a = 0

ALIGARH MUSLIM UNIVERSITY +2 ENTRANCE TEST PAPER

Answer 1

HELLO DEAR,

given that:-

tan α. tan β = a =c/a

and

α + β = π/6

=> α + β = 30°

=> tan(α + β )= tan30°

=> tan(α + β )=1/√3

=>
$\frac{tan \alpha + tan \beta }{1 - tan \alpha .tan \beta } = \frac{1}{ \sqrt{3} } \\ \\ = > \frac{tan \alpha + tan \beta }{1 - a} = \frac{1}{ \sqrt{3} } \\ \\ = > tan \alpha + tan \beta = \frac{(1 - a)}{ \sqrt{3} } = \frac{ - b}{a}$

we know that the formula Quadratic Equation:-

=> x² - (tanα + tanβ )x + (tanα.tanβ)=0

=> x² - [(1-a)/√3]x + a=0

=> √3x²- (1-a)x +√3a =0

HENCE OPTION (A) IS CORRECT

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

HEYA mate!!

Your solution is in the attachment.

Hope it helps you :-/