What is the general solution for tan^2(γ)+1=cos^2(γ) ?

The general solution for tan^2(γ)+1=cos^2(γ) is γ=2pin,γ=pi+2pin

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tan^2(γ)+1=cos^2(γ)

Solution

tan^{2} (γ) + 1 = cos^{2} (γ)

Solution

γ = 2 πn, γ = π + 2 πn

Degrees

γ = 0^{\circ} + 36 0^{\circ} n, γ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

tan^{2} (γ) + 1 = cos^{2} (γ)

Subtract

cos^{2} (γ)

from both sides

tan^{2} (γ) + 1 - cos^{2} (γ) = 0

Rewrite using trig identities

1 - cos^{2} (γ) + tan^{2} (γ)

Use the Pythagorean identity:

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

= - cos^{2} (γ) + sec^{2} (γ)

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

= - (\frac{1}{sec ( γ )})^{2} + sec^{2} (γ)

(\frac{1}{sec ( γ )})^{2} = \frac{1}{sec ^{2} ( γ )}

(\frac{1}{sec ( γ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( γ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( γ )}

= - \frac{1}{sec ^{2} ( γ )} + sec^{2} (γ)

- \frac{1}{sec ^{2} ( γ )} + sec^{2} (γ) = 0

Solve by substitution

- \frac{1}{sec ^{2} ( γ )} + sec^{2} (γ) = 0

Let:

sec (γ) = u

- \frac{1}{u ^{2}} + u^{2} = 0

- \frac{1}{u ^{2}} + u^{2} = 0 : u = 1, u = - 1, u = i, u = - i

- \frac{1}{u ^{2}} + u^{2} = 0

Multiply both sides by

u^{2}

- \frac{1}{u ^{2}} + u^{2} = 0

Multiply both sides by

u^{2}

- \frac{1}{u ^{2}} u^{2} + u^{2} u^{2} = 0 \cdot u^{2}

Simplify

- \frac{1}{u ^{2}} u^{2} + u^{2} u^{2} = 0 \cdot u^{2}

Simplify

- \frac{1}{u ^{2}} u^{2} : - 1

- \frac{1}{u ^{2}} u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u ^{2}}{u ^{2}}

Cancel the common factor:

u^{2}

= - 1

Simplify

u^{2} u^{2} : u^{4}

u^{2} u^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

Simplify

0 \cdot u^{2} : 0

0 \cdot u^{2}

Apply rule

0 \cdot a = 0

= 0

- 1 + u^{4} = 0

- 1 + u^{4} = 0

- 1 + u^{4} = 0

Solve

- 1 + u^{4} = 0 : u = 1, u = - 1, u = i, u = - i

- 1 + u^{4} = 0

Move

1

to the right side

- 1 + u^{4} = 0

Add

1

to both sides

- 1 + u^{4} + 1 = 0 + 1

Simplify

u^{4} = 1

u^{4} = 1

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} = 1

Solve

v^{2} = 1 : v = 1, v = - 1

v^{2} = 1

For

(g (x))^{2} = f (a)

the solutions are

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

v = 1, v = - 1

v = 1, v = - 1

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 1

Apply rule

1 = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Solve

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

Apply rule

1 = 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

The solutions are

u = 1, u = - 1, u = i, u = - i

u = 1, u = - 1, u = i, u = - i

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- \frac{1}{u ^{2}} + u^{2}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - 1, u = i, u = - i

Substitute back

u = sec (γ)

sec (γ) = 1, sec (γ) = - 1, sec (γ) = i, sec (γ) = - i

sec (γ) = 1, sec (γ) = - 1, sec (γ) = i, sec (γ) = - i

sec (γ) = 1 : γ = 2 πn

sec (γ) = 1

General solutions for

sec (γ) = 1

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

γ = 0 + 2 πn

γ = 0 + 2 πn

Solve

γ = 0 + 2 πn : γ = 2 πn

γ = 0 + 2 πn

0 + 2 πn = 2 πn

γ = 2 πn

γ = 2 πn

sec (γ) = - 1 : γ = π + 2 πn

sec (γ) = - 1

General solutions for

sec (γ) = - 1

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

γ = π + 2 πn

γ = π + 2 πn

sec (γ) = i :

No Solution

sec (γ) = i

No Solution

sec (γ) = - i :

No Solution

sec (γ) = - i

No Solution

Combine all the solutions

γ = 2 πn, γ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(γ)+1=cos^2(γ) ?
The general solution for tan^2(γ)+1=cos^2(γ) is γ=2pin,γ=pi+2pin