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

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

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Popular >

1/((1+tan(a)))=cot(a)

Solution

\frac{1}{( 1 + tan ( a ) )} = cot (a)

Solution

No Solution for a \in R

Solution steps

\frac{1}{( 1 + tan ( a ) )} = cot (a)

Subtract

cot (a)

from both sides

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

Simplify

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

\frac{1}{1 + tan ( a )} - cot (a)

Convert element to fraction:

cot (a) = \frac{cot ( a ) ( 1 + tan ( a ) )}{1 + tan ( a )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

1 - cot (a) (1 + tan (a)) = 0

Rewrite using trig identities

1 - (1 + tan (a)) cot (a)

Use the basic trigonometric identity:

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

= 1 - (1 + \frac{1}{cot ( a )}) cot (a)

Simplify

1 - (1 + \frac{1}{cot ( a )}) cot (a) : - cot (a)

1 - (1 + \frac{1}{cot ( a )}) cot (a)

= 1 - cot (a) (1 + \frac{1}{cot ( a )})

Expand

- cot (a) (1 + \frac{1}{cot ( a )}) : - cot (a) - 1

- cot (a) (1 + \frac{1}{cot ( a )})

Apply the distributive law:

a (b + c) = ab + a c

a = - cot (a), b = 1, c = \frac{1}{cot ( a )}

= - cot (a) \cdot 1 + (- cot (a)) \frac{1}{cot ( a )}

Apply minus-plus rules

+ (- a) = - a

= - 1 \cdot cot (a) - \frac{1}{cot ( a )} cot (a)

Simplify

- 1 \cdot cot (a) - \frac{1}{cot ( a )} cot (a) : - cot (a) - 1

- 1 \cdot cot (a) - \frac{1}{cot ( a )} cot (a)

1 \cdot cot (a) = cot (a)

1 \cdot cot (a)

Multiply:

1 \cdot cot (a) = cot (a)

= cot (a)

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

\frac{1}{cot ( a )} cot (a)

Multiply fractions:

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

= \frac{1 \cdot cot ( a )}{cot ( a )}

Cancel the common factor:

cot (a)

= 1

= - cot (a) - 1

= - cot (a) - 1

= 1 - cot (a) - 1

Simplify

1 - cot (a) - 1 : - cot (a)

1 - cot (a) - 1

Group like terms

= - cot (a) + 1 - 1

1 - 1 = 0

= - cot (a)

= - cot (a)

= - cot (a)

- cot (a) = 0

Divide both sides by

- 1

- cot (a) = 0

Divide both sides by

- 1

\frac{- cot ( a )}{- 1} = \frac{0}{- 1}

Simplify

cot (a) = 0

cot (a) = 0

General solutions for

cot (a) = 0

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

a = \frac{π}{2} + πn

a = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

No Solution for a \in R

Popular Examples

tan(2x)-2=3tan(x)

tan (2 x) - 2 = 3 tan (x)

sin(x-pi/6)=0

sin (x - \frac{π}{6}) = 0

2= 1/(cos^2(x))

2 = \frac{1}{cos ^{2} ( x )}

arccot(x-2)=arccot(x-1)+arccot(x)

arccot (x - 2) = arccot (x - 1) + arccot (x)

tan(x)-2/(tan(x))=1

tan (x) - \frac{2}{tan ( x )} = 1

Frequently Asked Questions (FAQ)

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