What is the general solution for tan^2(β)=1 ?

The general solution for tan^2(β)=1 is β= pi/4+pin,β=(3pi)/4+pin

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tan^2(β)=1

tan^{2} (β) = 1

β = \frac{π}{4} + πn, β = \frac{3 π}{4} + πn

Degrees

β = 4 5^{\circ} + 18 0^{\circ} n, β = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (β) = 1

Solve by substitution

tan^{2} (β) = 1

Let:

tan (β) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (β)

tan (β) = 1, tan (β) = - 1

tan (β) = 1, tan (β) = - 1

tan (β) = 1 : β = \frac{π}{4} + πn

tan (β) = 1

General solutions for

tan (β) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

β = \frac{π}{4} + πn

β = \frac{π}{4} + πn

tan (β) = - 1 : β = \frac{3 π}{4} + πn

tan (β) = - 1

General solutions for

tan (β) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

β = \frac{3 π}{4} + πn

β = \frac{3 π}{4} + πn

Combine all the solutions

β = \frac{π}{4} + πn, β = \frac{3 π}{4} + πn

6 sin (C) + 8 = 0

cos (2 t) = - sin (t)

sin^{2} (x) - 5 cos (x) - 5 = 0

\frac{1 + tan ^{2} ( 27 \cdot θ )}{1 - tan ^{2} ( 27 \cdot θ )} = 2

tan (x) = - cot (x)

What is the general solution for tan^2(β)=1 ?
The general solution for tan^2(β)=1 is β= pi/4+pin,β=(3pi)/4+pin