What is the general solution for tan^2(x)+1=a^2 ?

The general solution for tan^2(x)+1=a^2 is x=arctan(sqrt(a^2-1))+pin,x=arctan(-sqrt(a^2-1))+pin

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tan^2(x)+1=a^2

tan^{2} (x) + 1 = a^{2}

x = arctan (a^{2} - 1) + πn, x = arctan (- a^{2} - 1) + πn

Solution steps

tan^{2} (x) + 1 = a^{2}

Solve by substitution

tan^{2} (x) + 1 = a^{2}

Let:

tan (x) = u

u^{2} + 1 = a^{2}

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

u^{2} + 1 = a^{2}

Move

1

to the right side

u^{2} + 1 = a^{2}

Subtract

1

from both sides

u^{2} + 1 - 1 = a^{2} - 1

Simplify

u^{2} = a^{2} - 1

u^{2} = a^{2} - 1

For

x^{2} = f (a)

the solutions are

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

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

Substitute back

u = tan (x)

tan (x) = a^{2} - 1, tan (x) = - a^{2} - 1

tan (x) = a^{2} - 1, tan (x) = - a^{2} - 1

tan (x) = a^{2} - 1 : x = arctan (a^{2} - 1) + πn

tan (x) = a^{2} - 1

Apply trig inverse properties

tan (x) = a^{2} - 1

General solutions for

tan (x) = a^{2} - 1

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (a^{2} - 1) + πn

x = arctan (a^{2} - 1) + πn

tan (x) = - a^{2} - 1 : x = arctan (- a^{2} - 1) + πn

tan (x) = - a^{2} - 1

Apply trig inverse properties

tan (x) = - a^{2} - 1

General solutions for

tan (x) = - a^{2} - 1

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (- a^{2} - 1) + πn

x = arctan (- a^{2} - 1) + πn

Combine all the solutions

x = arctan (a^{2} - 1) + πn, x = arctan (- a^{2} - 1) + πn

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50 sin (4 5^{\circ}) = 600 sin (θ)

a, tan (a \cdot x^{0.5}) = - x^{0.5}

sin (x) = 35, csc (x)

2 + 2 cos (x) = \frac{2}{1 + cos ( x )}

What is the general solution for tan^2(x)+1=a^2 ?
The general solution for tan^2(x)+1=a^2 is x=arctan(sqrt(a^2-1))+pin,x=arctan(-sqrt(a^2-1))+pin