Let α= tan -1(1)+(1/2) tan -1(2)+(1/3) tan -1(3) and β= cot -1(1)+2 cot -1(2)+3 cot -1(3), then

Question

Let α= tan -1(1)+(1/2) tan -1(2)+(1/3) tan -1(3) and β= cot -1(1)+2 cot -1(2)+3 cot -1(3), then (A) 6 α- cot -1 3=(13 π/4) (B) β- cot -1 3=(3 π/4) (C) 6 α-β=(5 π/2) (D) 6 α+β=4 π

Tardigrade Admin · Accepted Answer

6 α=6 ⋅ (π/4)+3 tan -1 2+2 tan -1 3 ∴ 6 α- cot -1 3=(3 π/2)+3 ⋅ (3 π/4)-(π/2)=π+(9 π/4)=(13 π/4) and β- cot -1 3=(π/4)+(2 π/4)=(3 π/4).

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

$6 α - cot^{- 1} 3 = \frac{13 π}{4}$

$β - cot^{- 1} 3 = \frac{3 π}{4}$

$6 α - β = \frac{5 π}{2}$

$6 α + β = 4 π$

$6 α = 6 \cdot \frac{π}{4} + 3 tan^{- 1} 2 + 2 tan^{- 1} 3$
$∴ 6 α - cot^{- 1} 3 = \frac{3 π}{2} + 3 \cdot \frac{3 π}{4} - \frac{π}{2} = π + \frac{9 π}{4} = \frac{13 π}{4}$
and $β - cot^{- 1} 3 = \frac{π}{4} + \frac{2 π}{4} = \frac{3 π}{4}$ .

Q. Let α=tan−1(1)+21​tan−1(2)+31​tan−1(3) and β=cot−1(1)+2cot−1(2)+3cot−1(3), then

6α−cot−13=413π​

β−cot−13=43π​

6α−β=25π​

6α+β=4π