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

The general solution for sec^2(b)=2+tan(b) is b=1.01722…+pin,b=-0.55357…+pin

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Popular Trigonometry >

sec^2(b)=2+tan(b)

Solution

sec^{2} (b) = 2 + tan (b)

Solution

b = 1.01722 \dots + πn, b = - 0.55357 \dots + πn

Degrees

b = 58.28252 \dots^{\circ} + 18 0^{\circ} n, b = - 31.71747 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sec^{2} (b) = 2 + tan (b)

Subtract

2 + tan (b)

from both sides

sec^{2} (b) - 2 - tan (b) = 0

Rewrite using trig identities

- 2 + sec^{2} (b) - tan (b)

Use the Pythagorean identity:

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

= - 2 + tan^{2} (b) + 1 - tan (b)

Simplify

- 2 + tan^{2} (b) + 1 - tan (b) : tan^{2} (b) - tan (b) - 1

- 2 + tan^{2} (b) + 1 - tan (b)

Group like terms

= tan^{2} (b) - tan (b) - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

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

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

- 1 - tan (b) + tan^{2} (b) = 0

Solve by substitution

- 1 - tan (b) + tan^{2} (b) = 0

Let:

tan (b) = u

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

- 1 - u + u^{2} = 0 : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

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

Write in the standard form

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

u^{2} - u - 1 = 0

Solve with the quadratic formula

u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = - 1

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

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 1) = 5

(- 1)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

u = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

\frac{- ( - 1 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

Substitute back

u = tan (b)

tan (b) = \frac{1 + 5}{2}, tan (b) = \frac{1 - 5}{2}

tan (b) = \frac{1 + 5}{2}, tan (b) = \frac{1 - 5}{2}

tan (b) = \frac{1 + 5}{2} : b = arctan (\frac{1 + 5}{2}) + πn

tan (b) = \frac{1 + 5}{2}

Apply trig inverse properties

tan (b) = \frac{1 + 5}{2}

General solutions for

tan (b) = \frac{1 + 5}{2}

tan (x) = a \Rightarrow x = arctan (a) + πn

b = arctan (\frac{1 + 5}{2}) + πn

b = arctan (\frac{1 + 5}{2}) + πn

tan (b) = \frac{1 - 5}{2} : b = arctan (\frac{1 - 5}{2}) + πn

tan (b) = \frac{1 - 5}{2}

Apply trig inverse properties

tan (b) = \frac{1 - 5}{2}

General solutions for

tan (b) = \frac{1 - 5}{2}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

b = arctan (\frac{1 - 5}{2}) + πn

b = arctan (\frac{1 - 5}{2}) + πn

Combine all the solutions

b = arctan (\frac{1 + 5}{2}) + πn, b = arctan (\frac{1 - 5}{2}) + πn

Show solutions in decimal form

b = 1.01722 \dots + πn, b = - 0.55357 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(b)=2+tan(b) ?
The general solution for sec^2(b)=2+tan(b) is b=1.01722…+pin,b=-0.55357…+pin