Let α=2 tan -1 (1/2)+ sin -1 (3/5) and β= sin -1 (12/13)+ cos -1 (4/5)+ cot -1 (16/63) be such that 2 sin α and cos β are roots of the equation x2-p x+q=0, then find (p-q).

Question

Let α=2 tan -1 (1/2)+ sin -1 (3/5) and β= sin -1 (12/13)+ cos -1 (4/5)+ cot -1 (16/63) be such that 2 sin α and cos β are roots of the equation x2-p x+q=0, then find (p-q). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

α=(π/2), β=π ∴ 2 sin α=2 text and cos β=-1 text So, p =1, q=-2

Q. Let $α = 2 tan^{- 1} \frac{1}{2} + sin^{- 1} \frac{3}{5}$ and $β = sin^{- 1} \frac{12}{13} + cos^{- 1} \frac{4}{5} + cot^{- 1} \frac{16}{63}$ be such that $2 sin α$ and $cos β$ are roots of the equation $x^{2} - p x + q = 0$ , then find $(p - q)$ .

$α = \frac{π}{2}, β = π$
$∴ 2 sin α = 2 and cos β = - 1$
$So, p = 1, q = - 2$

Q. Let α=2tan−121​+sin−153​ and β=sin−11312​+cos−154​+cot−16316​ be such that 2sinα and cosβ are roots of the equation x2−px+q=0, then find (p−q).

α=2π​,β=π ∴2sinα=2 and cosβ=−1 So, p =1,q=−2

Q. Let $α = 2 tan^{- 1} \frac{1}{2} + sin^{- 1} \frac{3}{5}$ and $β = sin^{- 1} \frac{12}{13} + cos^{- 1} \frac{4}{5} + cot^{- 1} \frac{16}{63}$ be such that $2 sin α$ and $cos β$ are roots of the equation $x^{2} - p x + q = 0$ , then find $(p - q)$ .

$α = \frac{π}{2}, β = π$
$∴ 2 sin α = 2 and cos β = - 1$
$So, p = 1, q = - 2$