Is cot(x)=(1+cot(x))/(1+tan(x)) ?

The answer to whether cot(x)=(1+cot(x))/(1+tan(x)) is True

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prove cot(x)=(1+cot(x))/(1+tan(x))

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 + tan ( x )}

Use the basic trigonometric identity:

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

= \frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}}

Simplify

\frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}} : \frac{cos ( x )}{sin ( x )}

\frac{1 + \frac{c o s ( x )}{s i n ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}}

Join

1 + \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) + sin ( x )}{cos ( x )}

1 + \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

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

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot cos ( x ) + sin ( x )}{cos ( x )}

Multiply:

1 \cdot cos (x) = cos (x)

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

= \frac{1 + \frac{c o s ( x )}{s i n ( x )}}{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}

Join

1 + \frac{cos ( x )}{sin ( x )} : \frac{sin ( x ) + cos ( x )}{sin ( x )}

1 + \frac{cos ( x )}{sin ( x )}

Convert element to fraction:

1 = \frac{1 sin ( x )}{sin ( x )}

= \frac{1 \cdot sin ( x )}{sin ( x )} + \frac{cos ( x )}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sin ( x ) + cos ( x )}{sin ( x )}

Multiply:

1 \cdot sin (x) = sin (x)

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

= \frac{\frac{s i n ( x ) + c o s ( x )}{s i n ( x )}}{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}

Divide fractions:

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

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

Cancel the common factor:

sin (x) + cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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

6 cos^{2} (B) - 6 sin^{2} (B) = 12 cos^{2} (B) - 6

cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

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

\frac{sec ( θ ) cos ( θ )}{tan ( θ )} = cot (θ)

Is cot(x)=(1+cot(x))/(1+tan(x)) ?
The answer to whether cot(x)=(1+cot(x))/(1+tan(x)) is True