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

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

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

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

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sec^{2} (x)

from both sides

1 - tan^{2} (x) - sec^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - 2 tan^{2} (x)

- 2 tan^{2} (x) = 0

Divide both sides by

- 2

- 2 tan^{2} (x) = 0

Divide both sides by

- 2

- 2 tan^{2} (x) = 0

Divide both sides by

- 2

\frac{- 2 tan ^{2} ( x )}{- 2} = \frac{0}{- 2}

Simplify

tan^{2} (x) = 0

tan^{2} (x) = 0

Apply rule

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

tan (x) = 0

General solutions for

tan (x) = 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}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

cos (2 θ - \frac{π}{2}) = - 1, 0 \leq θ < 2 π

sinh (x) = - \frac{3}{4}

tan (3 θ) = 1

7 - 7 sin (θ) = 7

- 12 cos^{2} (θ) - 8 = - 17

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