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

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

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Popular >

sec(θ)=tan(θ)+cot(θ)

Solution

sec (θ) = tan (θ) + cot (θ)

Solution

No Solution for θ \in R

Solution steps

sec (θ) = tan (θ) + cot (θ)

Subtract

tan (θ) + cot (θ)

from both sides

sec (θ) - tan (θ) - cot (θ) = 0

Express with sin, cos

- cot (θ) + sec (θ) - tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

- \frac{cos ( θ )}{sin ( θ )} + \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 )}

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

Least Common Multiplier of

sin (θ), cos (θ) : sin (θ) cos (θ)

sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

cos (θ)

= sin (θ) cos (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (θ) cos (θ)

For

\frac{cos ( θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

For

\frac{1 - sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Rewrite using trig identities

- cos^{2} (θ) + (1 - sin (θ)) sin (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

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

Expand

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

sin (θ) (1 - sin (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = sin (θ), b = 1, c = sin (θ)

= sin (θ) \cdot 1 - sin (θ) sin (θ)

= 1 \cdot sin (θ) - sin (θ) sin (θ)

Simplify

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

1 \cdot sin (θ) - sin (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= sin (θ) - sin^{2} (θ)

= sin (θ) - sin^{2} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

sin^{2} (θ) - sin^{2} (θ) = 0

= sin (θ) - 1

= sin (θ) - 1

= sin (θ) - 1

- 1 + sin (θ) = 0

Move

1

to the right side

- 1 + sin (θ) = 0

Add

1

to both sides

- 1 + sin (θ) + 1 = 0 + 1

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 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}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

No Solution for θ \in R

Popular Examples

sec^2(x)= 4/3 ,0<= x<= 2pi

sec^{2} (x) = \frac{4}{3}, 0 \leq x \leq 2 π

sqrt(3)tan(3θ)-1=0

3 tan (3 θ) - 1 = 0

2tan(x)=sec(x)

2 tan (x) = sec (x)

cos(x)=2pi

cos (x) = 2 π

sin(6x)+sin(2x)=0

sin (6 x) + sin (2 x) = 0

Frequently Asked Questions (FAQ)

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