What is the general solution for tan^2(θ)=1-sec^2(θ) ?

The general solution for tan^2(θ)=1-sec^2(θ) is θ=pin

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tan^2(θ)=1-sec^2(θ)

tan^{2} (θ) = 1 - sec^{2} (θ)

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) = 1 - sec^{2} (θ)

Subtract

1 - sec^{2} (θ)

from both sides

tan^{2} (θ) - 1 + sec^{2} (θ) = 0

Rewrite using trig identities

- 1 + sec^{2} (θ) + tan^{2} (θ)

Use the Pythagorean identity:

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

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

= tan^{2} (θ) + tan^{2} (θ)

Simplify

= 2 tan^{2} (θ)

2 tan^{2} (θ) = 0

Divide both sides by

2

2 tan^{2} (θ) = 0

Divide both sides by

2

2 tan^{2} (θ) = 0

Divide both sides by

2

\frac{2 tan ^{2} ( θ )}{2} = \frac{0}{2}

Simplify

tan^{2} (θ) = 0

tan^{2} (θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

tan (θ) = 0

General solutions for

tan (θ) = 0

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}

θ = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

x, sin (4 x) = 0

cot (θ) = - \frac{7}{8}

cos (2 x) = - sin (2 x)

α, sin^{2} (α) + sin^{2} (β) = 1

- 6 cos (A) + 8 = 3 cos (A) + 8

What is the general solution for tan^2(θ)=1-sec^2(θ) ?
The general solution for tan^2(θ)=1-sec^2(θ) is θ=pin