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

The general solution for tan^2(θ)=sec^2(θ)+1 is No Solution for θ\in\mathbb{R}

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tan^2(θ)=sec^2(θ)+1

Solution

tan^{2} (θ) = sec^{2} (θ) + 1

Solution

No Solution for θ \in R

Solution steps

tan^{2} (θ) = sec^{2} (θ) + 1

Subtract

sec^{2} (θ) + 1

from both sides

tan^{2} (θ) - sec^{2} (θ) - 1 = 0

Express with sin, cos

- 1 - sec^{2} (θ) + tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 - (\frac{1}{cos ( θ )})^{2} + (\frac{sin ( θ )}{cos ( θ )})^{2}

Simplify

- 1 - (\frac{1}{cos ( θ )})^{2} + (\frac{sin ( θ )}{cos ( θ )})^{2} : \frac{- cos ^{2} ( θ ) - 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

- 1 - (\frac{1}{cos ( θ )})^{2} + (\frac{sin ( θ )}{cos ( θ )})^{2}

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

(\frac{sin ( θ )}{cos ( θ )})^{2} = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

= - 1 - \frac{1}{cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Combine the fractions

- \frac{1}{cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} : \frac{- 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

Apply rule

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

= \frac{- 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

= - 1 + \frac{sin ^{2} ( θ ) - 1}{cos ^{2} ( θ )}

Convert element to fraction:

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

= - \frac{1 \cdot cos ^{2} ( θ )}{cos ^{2} ( θ )} + \frac{- 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ^{2} ( θ ) - 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

= \frac{- cos ^{2} ( θ ) - 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{- cos ^{2} ( θ ) - 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

\frac{- 1 - cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 - cos^{2} (θ) + sin^{2} (θ) = 0

Rewrite using trig identities

- 1 - cos^{2} (θ) + sin^{2} (θ)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

- cos^{2} (x) + sin^{2} (x) = - cos (2 x)

= - 1 - cos (2 θ)

- 1 - cos (2 θ) = 0

Move

1

to the right side

- 1 - cos (2 θ) = 0

Add

1

to both sides

- 1 - cos (2 θ) + 1 = 0 + 1

Simplify

- cos (2 θ) = 1

- cos (2 θ) = 1

Divide both sides by

- 1

- cos (2 θ) = 1

Divide both sides by

- 1

\frac{- cos ( 2 θ )}{- 1} = \frac{1}{- 1}

Simplify

cos (2 θ) = - 1

cos (2 θ) = - 1

General solutions for

cos (2 θ) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 θ = π + 2 πn

2 θ = π + 2 πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(θ)=sec^2(θ)+1 ?
The general solution for tan^2(θ)=sec^2(θ)+1 is No Solution for θ\in\mathbb{R}