Let -(π/6)β2, then α1+β2 equals

Question

Let -(π/6)<θ<-(π/12). Suppose α1 and β1 are the roots of the equation x2-2 x sec θ+1=0 and α2 and β2 are the roots of the equation x 2+2 x tan θ-1=0. If α1>β1 and α2>β2, then α1+β2 equals (A) 2( sec θ- tan θ) (B) 2 sec θ (C) -2 tan θ (D) 0

Tardigrade Admin · Accepted Answer

x2-2 x sec θ+1=0 Now roots are x = sec θ ± tan θ So α1= sec θ- tan θ β1= sec θ+ tan θ begincases text as θ ∈(-(π/6),-(π/12)) α1>β1 endcases x2+2 x tan θ-1=0 Now roots are x =± sec θ- tan θ So,α2=+ sec θ- tan θ β2=- sec θ- tan θ ∴ α1+β2=-2 tan θ

Q. Let $- \frac{π}{6} < θ < - \frac{π}{12}$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

$2 (sec θ - tan θ)$

$2 sec θ$

$- 2 tan θ$

$0$

Q. Let −6π​<θ<−12π​. Suppose α1​ and β1​ are the roots of the equation x2−2xsecθ+1=0 and α2​ and β2​ are the roots of the equation x2+2xtanθ−1=0. If α1​>β1​ and α2​>β2​, then α1​+β2​ equals

2(secθ−tanθ)

2secθ

−2tanθ

0

x2−2xsecθ+1=0 Now roots are x=secθ±tanθ So α1​=secθ−tanθ & β1​=secθ+tanθ { as θ∈(−6π​,−12π​)&α1​>β1​​ &x2+2xtanθ−1=0 Now roots are x=±secθ−tanθ So,α2​=+secθ−tanθ β2​=−secθ−tanθ ∴α1​+β2​=−2tanθ

Q. Let $- \frac{π}{6} < θ < - \frac{π}{12}$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x sec θ + 1 = 0$ and $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x tan θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

$2 (sec θ - tan θ)$

$2 sec θ$

$- 2 tan θ$

$0$