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

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

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Popular Trigonometry >

sec(a)=1+tan(a)

Solution

sec (a) = 1 + tan (a)

Solution

a = 2 πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n

Solution steps

sec (a) = 1 + tan (a)

Subtract

1 + tan (a)

from both sides

sec (a) - 1 - tan (a) = 0

Express with sin, cos

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

Simplify

\frac{1}{cos ( a )} - 1 - \frac{sin ( a )}{cos ( a )} : \frac{1 - sin ( a ) - cos ( a )}{cos ( a )}

\frac{1}{cos ( a )} - 1 - \frac{sin ( a )}{cos ( a )}

Combine the fractions

\frac{1}{cos ( a )} - \frac{sin ( a )}{cos ( a )} : \frac{1 - sin ( a )}{cos ( a )}

Apply rule

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

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

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

Convert element to fraction:

1 = \frac{1 cos ( a )}{cos ( a )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (a) = cos (a)

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

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

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

1 - sin (a) - cos (a) = 0

Add

cos (a)

to both sides

1 - sin (a) = cos (a)

Square both sides

(1 - sin (a))^{2} = cos^{2} (a)

Subtract

cos^{2} (a)

from both sides

(1 - sin (a))^{2} - cos^{2} (a) = 0

Rewrite using trig identities

(1 - sin (a))^{2} - cos^{2} (a)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= (1 - sin (a))^{2} - (1 - sin^{2} (a))

Simplify

(1 - sin (a))^{2} - (1 - sin^{2} (a)) : 2 sin^{2} (a) - 2 sin (a)

(1 - sin (a))^{2} - (1 - sin^{2} (a))

(1 - sin (a))^{2} : 1 - 2 sin (a) + sin^{2} (a)

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = sin (a)

= 1^{2} - 2 \cdot 1 \cdot sin (a) + sin^{2} (a)

Simplify

1^{2} - 2 \cdot 1 \cdot sin (a) + sin^{2} (a) : 1 - 2 sin (a) + sin^{2} (a)

1^{2} - 2 \cdot 1 \cdot sin (a) + sin^{2} (a)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot sin (a) + sin^{2} (a)

Multiply the numbers:

2 \cdot 1 = 2

= 1 - 2 sin (a) + sin^{2} (a)

= 1 - 2 sin (a) + sin^{2} (a)

= 1 - 2 sin (a) + sin^{2} (a) - (1 - sin^{2} (a))

- (1 - sin^{2} (a)) : - 1 + sin^{2} (a)

- (1 - sin^{2} (a))

Distribute parentheses

= - (1) - (- sin^{2} (a))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (a)

= 1 - 2 sin (a) + sin^{2} (a) - 1 + sin^{2} (a)

Simplify

1 - 2 sin (a) + sin^{2} (a) - 1 + sin^{2} (a) : 2 sin^{2} (a) - 2 sin (a)

1 - 2 sin (a) + sin^{2} (a) - 1 + sin^{2} (a)

Group like terms

= - 2 sin (a) + sin^{2} (a) + sin^{2} (a) + 1 - 1

Add similar elements:

sin^{2} (a) + sin^{2} (a) = 2 sin^{2} (a)

= - 2 sin (a) + 2 sin^{2} (a) + 1 - 1

1 - 1 = 0

= 2 sin^{2} (a) - 2 sin (a)

= 2 sin^{2} (a) - 2 sin (a)

= 2 sin^{2} (a) - 2 sin (a)

- 2 sin (a) + 2 sin^{2} (a) = 0

Solve by substitution

- 2 sin (a) + 2 sin^{2} (a) = 0

Let:

sin (a) = u

- 2 u + 2 u^{2} = 0

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

- 2 u + 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} - 2 u = 0

Solve with the quadratic formula

2 u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 2 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

u_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 2}, u_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 2}

u = \frac{- ( - 2 ) + 2}{2 \cdot 2} : 1

\frac{- ( - 2 ) + 2}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{2 + 2}{2 \cdot 2}

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{2 - 2}{2 \cdot 2}

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = sin (a)

sin (a) = 1, sin (a) = 0

sin (a) = 1, sin (a) = 0

sin (a) = 1 : a = \frac{π}{2} + 2 πn

sin (a) = 1

General solutions for

sin (a) = 1

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}

a = \frac{π}{2} + 2 πn

a = \frac{π}{2} + 2 πn

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 0

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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

Combine all the solutions

a = \frac{π}{2} + 2 πn, a = 2 πn, a = π + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

sec (a) = 1 + tan (a)

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

sec (a) = 1 + tan (a)

plug in

a = \frac{π}{2} + 2 π 1

sec (\frac{π}{2} + 2 π 1) = 1 + tan (\frac{π}{2} + 2 π 1)

Refine

\infty = \infty

\Rightarrow True

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

sec (a) = 1 + tan (a)

plug in

a = 2 π 1

sec (2 π 1) = 1 + tan (2 π 1)

Refine

1 = 1

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

sec (a) = 1 + tan (a)

plug in

a = π + 2 π 1

sec (π + 2 π 1) = 1 + tan (π + 2 π 1)

Refine

- 1 = 1

\Rightarrow False

a = \frac{π}{2} + 2 πn, a = 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

a = 2 πn

Graph

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Frequently Asked Questions (FAQ)

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