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Popular Trigonometry >

solvefor x,tan(x)+cot(x)= 1/f

Solution

solve for

x, tan (x) + cot (x) = \frac{1}{f}

Solution

x = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn, x = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

Solution steps

tan (x) + cot (x) = \frac{1}{f}

Subtract

\frac{1}{f}

from both sides

tan (x) + cot (x) - \frac{1}{f} = 0

Simplify

tan (x) + cot (x) - \frac{1}{f} : \frac{f tan ( x ) + f cot ( x ) - 1}{f}

tan (x) + cot (x) - \frac{1}{f}

Convert element to fraction:

tan (x) = \frac{tan ( x ) f}{f}, cot (x) = \frac{cot ( x ) f}{f}

= \frac{tan ( x ) f}{f} + \frac{cot ( x ) f}{f} - \frac{1}{f}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( x ) f + cot ( x ) f - 1}{f}

\frac{f tan ( x ) + f cot ( x ) - 1}{f} = 0

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

f tan (x) + f cot (x) - 1 = 0

Rewrite using trig identities

- 1 + cot (x) f + tan (x) f

Use the basic trigonometric identity:

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

= - 1 + cot (x) f + \frac{1}{cot ( x )} f

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot f = f

= \frac{f}{cot ( x )}

= - 1 + f cot (x) + \frac{f}{cot ( x )}

- 1 + \frac{f}{cot ( x )} + cot (x) f = 0

Solve by substitution

- 1 + \frac{f}{cot ( x )} + cot (x) f = 0

Let:

cot (x) = u

- 1 + \frac{f}{u} + u f = 0

- 1 + \frac{f}{u} + u f = 0 : u = \frac{1 + 1 - 4 f ^{2}}{2 f}, u = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

- 1 + \frac{f}{u} + u f = 0

Multiply both sides by

u

- 1 + \frac{f}{u} + u f = 0

Multiply both sides by

u

- 1 \cdot u + \frac{f}{u} u + u f u = 0 \cdot u

Simplify

- 1 \cdot u + \frac{f}{u} u + u f u = 0 \cdot u

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - u

Simplify

\frac{f}{u} u : f

\frac{f}{u} u

Multiply fractions:

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

= \frac{f u}{u}

Cancel the common factor:

u

= f

Simplify

u f u : f u^{2}

u f u

Apply exponent rule:

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

uu = u^{1 + 1}

= f u^{1 + 1}

Add the numbers:

1 + 1 = 2

= f u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

- u + f + f u^{2} = 0 : u = \frac{1 + 1 - 4 f ^{2}}{2 f}, u = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 1)^{2} - 4 ff : 1 - 4 f^{2}

(- 1)^{2} - 4 ff

(- 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 ff = 4 f^{2}

4 ff

Apply exponent rule:

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

ff = f^{1 + 1}

= 4 f^{1 + 1}

Add the numbers:

1 + 1 = 2

= 4 f^{2}

= 1 - 4 f^{2}

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

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1 - 4 f ^{2}}{2 f}, u_{2} = \frac{- ( - 1 ) - 1 - 4 f ^{2}}{2 f}

u = \frac{- ( - 1 ) + 1 - 4 f ^{2}}{2 f} : \frac{1 + 1 - 4 f ^{2}}{2 f}

\frac{- ( - 1 ) + 1 - 4 f ^{2}}{2 f}

Apply rule

- (- a) = a

= \frac{1 + 1 - 4 f ^{2}}{2 f}

u = \frac{- ( - 1 ) - 1 - 4 f ^{2}}{2 f} : \frac{1 - 1 - 4 f ^{2}}{2 f}

\frac{- ( - 1 ) - 1 - 4 f ^{2}}{2 f}

Apply rule

- (- a) = a

= \frac{1 - 1 - 4 f ^{2}}{2 f}

The solutions to the quadratic equation are:

u = \frac{1 + 1 - 4 f ^{2}}{2 f}, u = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

u = \frac{1 + 1 - 4 f ^{2}}{2 f}, u = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

Substitute back

u = cot (x)

cot (x) = \frac{1 + 1 - 4 f ^{2}}{2 f}, cot (x) = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

cot (x) = \frac{1 + 1 - 4 f ^{2}}{2 f}, cot (x) = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

cot (x) = \frac{1 + 1 - 4 f ^{2}}{2 f} : x = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

cot (x) = \frac{1 + 1 - 4 f ^{2}}{2 f}

Apply trig inverse properties

cot (x) = \frac{1 + 1 - 4 f ^{2}}{2 f}

General solutions for

cot (x) = \frac{1 + 1 - 4 f ^{2}}{2 f}

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

x = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

cot (x) = \frac{1 - 1 - 4 f ^{2}}{2 f} : x = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

cot (x) = \frac{1 - 1 - 4 f ^{2}}{2 f}

Apply trig inverse properties

cot (x) = \frac{1 - 1 - 4 f ^{2}}{2 f}

General solutions for

cot (x) = \frac{1 - 1 - 4 f ^{2}}{2 f}

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

x = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

Combine all the solutions

x = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn, x = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

Graph

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