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

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

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1-tan^2(x)=(cos^2(x))/(cos^2(x))

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

x = πn

Degrees

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

Solution steps

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

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

\frac{cos ^{2} ( x )}{cos ^{2} ( x )}

Apply rule

\frac{a}{a} = 1

= 1

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

- tan^{2} (x) = 0

- tan^{2} (x) = 0

Divide both sides by

- 1

- tan^{2} (x) = 0

Divide both sides by

- 1

- tan^{2} (x) = 0

Divide both sides by

- 1

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

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

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

tan^{2} (y) + 1 = cos^{2} (y)

4 sin (2 x) + csc (2 x) = 4

cot (θ) = \frac{9}{40}

sin (a) = - \frac{1}{2}

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