What is the general solution for (cot(x)+1/(cot(x)))/(cos(x))=sec(x) ?

The general solution for (cot(x)+1/(cot(x)))/(cos(x))=sec(x) is No Solution for x\in\mathbb{R}

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(cot(x)+1/(cot(x)))/(cos(x))=sec(x)

Solution

\frac{cot ( x ) + \frac{1}{c o t ( x )}}{cos ( x )} = sec (x)

Solution

No Solution for x \in R

Solution steps

\frac{cot ( x ) + \frac{1}{c o t ( x )}}{cos ( x )} = sec (x)

Subtract

sec (x)

from both sides

\frac{cot ^{2} ( x ) + 1}{cot ( x ) cos ( x )} - sec (x) = 0

Simplify

\frac{cot ^{2} ( x ) + 1}{cot ( x ) cos ( x )} - sec (x) : \frac{cot ^{2} ( x ) + 1 - sec ( x ) cot ( x ) cos ( x )}{cot ( x ) cos ( x )}

\frac{cot ^{2} ( x ) + 1}{cot ( x ) cos ( x )} - sec (x)

Convert element to fraction:

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

= \frac{cot ^{2} ( x ) + 1}{cot ( x ) cos ( x )} - \frac{sec ( x ) cot ( x ) cos ( x )}{cot ( x ) cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{cot ^{2} ( x ) + 1 - sec ( x ) cot ( x ) cos ( x )}{cot ( x ) cos ( x )}

\frac{cot ^{2} ( x ) + 1 - sec ( x ) cot ( x ) cos ( x )}{cot ( x ) cos ( x )} = 0

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

cot^{2} (x) + 1 - sec (x) cot (x) cos (x) = 0

Rewrite using trig identities

1 + cot^{2} (x) - cos (x) cot (x) sec (x)

Use the basic trigonometric identity:

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

= 1 + cot^{2} (x) - \frac{1}{sec ( x )} cot (x) sec (x)

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

\frac{1}{sec ( x )} cot (x) sec (x)

Multiply fractions:

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

= \frac{1 \cdot cot ( x ) sec ( x )}{sec ( x )}

Cancel the common factor:

sec (x)

= 1 \cdot cot (x)

Multiply:

1 \cdot cot (x) = cot (x)

= cot (x)

= 1 + cot^{2} (x) - cot (x)

1 - cot (x) + cot^{2} (x) = 0

Solve by substitution

1 - cot (x) + cot^{2} (x) = 0

Let:

cot (x) = u

1 - u + u^{2} = 0

1 - u + u^{2} = 0 : u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{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}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 1 : 3 i

(- 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

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

u_{1, 2} = \frac{- ( - 1 ) \pm 3 i}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3 i}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 3 i}{2 \cdot 1}

u = \frac{- ( - 1 ) + 3 i}{2 \cdot 1} : \frac{1}{2} + i \frac{3}{2}

\frac{- ( - 1 ) + 3 i}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 + 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 3 i}{2}

Rewrite

\frac{1 + 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

\frac{1 + 3 i}{2}

Apply the fraction rule:

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

\frac{1 + 3 i}{2} = \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

u = \frac{- ( - 1 ) - 3 i}{2 \cdot 1} : \frac{1}{2} - i \frac{3}{2}

\frac{- ( - 1 ) - 3 i}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{1 - 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 3 i}{2}

Rewrite

\frac{1 - 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

\frac{1 - 3 i}{2}

Apply the fraction rule:

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

\frac{1 - 3 i}{2} = \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

The solutions to the quadratic equation are:

u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

Substitute back

u = cot (x)

cot (x) = \frac{1}{2} + i \frac{3}{2}, cot (x) = \frac{1}{2} - i \frac{3}{2}

cot (x) = \frac{1}{2} + i \frac{3}{2}, cot (x) = \frac{1}{2} - i \frac{3}{2}

cot (x) = \frac{1}{2} + i \frac{3}{2} :

No Solution

cot (x) = \frac{1}{2} + i \frac{3}{2}

No Solution

cot (x) = \frac{1}{2} - i \frac{3}{2} :

No Solution

cot (x) = \frac{1}{2} - i \frac{3}{2}

No Solution

Combine all the solutions

No Solution for x \in R

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(1 + csc (A)) (1 - csc (A)) = - cot (A)

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Frequently Asked Questions (FAQ)

What is the general solution for (cot(x)+1/(cot(x)))/(cos(x))=sec(x) ?
The general solution for (cot(x)+1/(cot(x)))/(cos(x))=sec(x) is No Solution for x\in\mathbb{R}