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

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

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1-sec^2(θ)=tan^2(θ)

1 - sec^{2} (θ) = tan^{2} (θ)

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

1 - sec^{2} (θ) = tan^{2} (θ)

Subtract

tan^{2} (θ)

from both sides

1 - sec^{2} (θ) - tan^{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

6 tan^{2} (x) = 5 sec (x)

x, cos^{2} (x) = 0.5

sin (x) = 0.24, 0 \leq x < 2 π

sec^{2} (θ) cot (θ) = 8

4 + 5 sin (x) = 1

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