Is ((1+tan(θ)))/((1+cot(θ)))=tan(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 + tan ( θ )}{1 + cot ( θ )}

Express with sin, cos

\frac{1 + tan ( θ )}{1 + cot ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 + \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Multiply:

1 \cdot sin (θ) = sin (θ)

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

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

Join

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

1 + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (θ) + sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{1}{sin ^{2} ( θ )} = 1 - tan^{2} (θ)

p ro v e \frac{1}{( sec ( θ ) ) ^{2}} + \frac{1}{( csc ( θ ) ) ^{2}} = 1

p ro v e sin^{2} (\frac{a}{2}) = \frac{1 - cos ( a )}{2}

p ro v e cot (x) + tan (x) = sec (csc (x))

p ro v e cos (\frac{1}{3} π) = cos (- \frac{1}{3} π)

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