Question

Let θ € (-π/6, -π/12) If a1 and b1 are roots of x²-2xsecθ+1=0. And if a2 and b2 are roots of x²+2xtanθ-1=0. If a1 is greater than b1 and a2 is greater than b2, then find the value of a1+b2.

Please give the answer with appropriate solution. I will mark them as brainliest and also give a thanks.​

Answer 1

Given info : a1 and b1 are roots of x² - 2xsecθ + 1 = 0 and a2 and b2 are roots of x² + 2xtanθ - 1 = 0. where a1 > b1 and a2 > b2 also θ ∈ (-π/6, -π/12)

To find : the value of a1 + b2 is ..

solution : case 1 : x² - 2xsecθ + 1 = 0

x = {2secθ ± √(4sec²θ - 4)}/2

= secθ ± tanθ [ ∵ sec²x - 1 = tan²x ]

here, a1 and b1 are roots where a1 > b1

as θ belongs to (-π/6, -π/12)

⇒ secθ - tanθ > secθ + tanθ

so, a1 = secθ - tanθ and b1 = secθ + tanθ

case 2 : x² + 2xtanθ - 1 = 0

x = {-2tanθ ± √(4tan²θ + 4)}/2

= -tanθ ± secθ [ ∵ tan²x + 1 = sec²x ]

here a2 and b2 are roots where a2 > b2

as θ belongs to (-π/6, -π/12)

⇒-tanθ + secθ > -tanθ - secθ

so, a2 = -tanθ + secθ and b2 = -tanθ - secθ

now the value of a1 + b2

= secθ - tanθ + ( -tanθ - secθ)

= -2tanθ

Therefore the value of a1 + b2 is -2tanθ

Answer 2

Step-by-step explanation:

above answer is correct