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

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

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Popular >

tan(θ)+1=sec(θ)

Solution

tan (θ) + 1 = sec (θ)

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (θ) + 1 = sec (θ)

Subtract

sec (θ)

from both sides

tan (θ) + 1 - sec (θ) = 0

Express with sin, cos

1 - sec (θ) + tan (θ)

Use the basic trigonometric identity:

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

= 1 - \frac{1}{cos ( θ )} + tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

= \frac{- 1 + sin ( θ )}{cos ( θ )}

= 1 + \frac{sin ( θ ) - 1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

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

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

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

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

- 1 + cos (θ) + sin (θ) = 0

Rewrite using trig identities

- 1 + cos (θ) + sin (θ)

sin (θ) + cos (θ) = 2 sin (θ + \frac{π}{4})

sin (θ) + cos (θ)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (θ) + sin (\frac{π}{4}) cos (θ))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (θ + \frac{π}{4})

= - 1 + 2 sin (θ + \frac{π}{4})

- 1 + 2 sin (θ + \frac{π}{4}) = 0

Move

1

to the right side

- 1 + 2 sin (θ + \frac{π}{4}) = 0

Add

1

to both sides

- 1 + 2 sin (θ + \frac{π}{4}) + 1 = 0 + 1

Simplify

2 sin (θ + \frac{π}{4}) = 1

2 sin (θ + \frac{π}{4}) = 1

Divide both sides by

2

2 sin (θ + \frac{π}{4}) = 1

Divide both sides by

2

\frac{2 sin ( θ + \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( θ + \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( θ + \frac{π}{4} )}{2} : sin (θ + \frac{π}{4})

\frac{2 sin ( θ + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (θ + \frac{π}{4})

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (θ + \frac{π}{4}) = \frac{2}{2}

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

θ + \frac{π}{4} = \frac{π}{4} + 2 πn, θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

θ + \frac{π}{4} = \frac{π}{4} + 2 πn, θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

θ + \frac{π}{4} = \frac{π}{4} + 2 πn : θ = 2 πn

θ + \frac{π}{4} = \frac{π}{4} + 2 πn

Subtract

\frac{π}{4}

from both sides

θ + \frac{π}{4} - \frac{π}{4} = \frac{π}{4} + 2 πn - \frac{π}{4}

Simplify

θ = 2 πn

Solve

θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn : θ = 2 πn + \frac{π}{2}

θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Move

\frac{π}{4}

to the right side

θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Subtract

\frac{π}{4}

from both sides

θ + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

Simplify

θ + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

Simplify

θ + \frac{π}{4} - \frac{π}{4} : θ

θ + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= θ

Simplify

\frac{3 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{2}

\frac{3 π}{4} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

Combine the fractions

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

Apply rule

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

= \frac{- π + 3 π}{4}

Add similar elements:

- π + 3 π = 2 π

= \frac{2 π}{4}

Cancel the common factor:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn, θ = 2 πn + \frac{π}{2}

Since the equation is undefined for:

2 πn + \frac{π}{2}

θ = 2 πn

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tan(x)+3=0

tan (x) + 3 = 0

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Frequently Asked Questions (FAQ)

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