If α=2 tan -1((1/2))+ tan -1(3) and β=3 sin -1((√3/2))+ sin -1((1/√2)), then

Question

If α=2 tan -1((1/2))+ tan -1(3) and β=3 sin -1((√3/2))+ sin -1((1/√2)), then (A) α>β (B) α=β (C) α ∈((π/2), (3 π/4)) (D) α ∈((3 π/4), π)

Tardigrade Admin · Accepted Answer

α=2 tan -1((1/2))+ tan -1 3= tan -1((2((1/2))/1-(1)4))+ tan -1 3= tan -1 (4/3)+ tan -1 3=π- tan -1((13/9)) ⇒ α ∈((π/2), (3 π/4)) and α<β, as β=3((π/3))+(π/4)=(5 π/4).

Q. If $α = 2 tan^{- 1} (\frac{1}{2}) + tan^{- 1} (3)$ and $β = 3 sin^{- 1} (\frac{3}{2}) + sin^{- 1} (\frac{1}{2})$ , then

$α > β$

$α = β$

$α \in (\frac{π}{2}, \frac{3 π}{4})$

$α \in (\frac{3 π}{4}, π)$

Q. If α=2tan−1(21​)+tan−1(3) and β=3sin−1(23​​)+sin−1(2​1​), then

α>β

α=β

α∈(2π​,43π​)

α∈(43π​,π)

α=2tan−1(21​)+tan−13=tan−1(1−41​2(21​)​)+tan−13=tan−134​+tan−13=π−tan−1(913​) ⇒α∈(2π​,43π​) and α<β, as β=3(3π​)+4π​=45π​.

Q. If $α = 2 tan^{- 1} (\frac{1}{2}) + tan^{- 1} (3)$ and $β = 3 sin^{- 1} (\frac{3}{2}) + sin^{- 1} (\frac{1}{2})$ , then

$α > β$

$α = β$

$α \in (\frac{π}{2}, \frac{3 π}{4})$

$α \in (\frac{3 π}{4}, π)$