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

Solution steps

tan^{2} (β) = 1

Solve by substitution

tan (β) = 1, tan (β) = - 1

tan (β) = 1 : β = \frac{π}{4} + πn

tan (β) = - 1 : 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