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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (x) + sin (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (θ) cot (θ) - sin^{2} (θ) = cos^{2} (θ)

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

sec (θ) = \frac{1}{cos ( θ )}

csc (2 θ) = \frac{csc ( θ )}{2 cos ( θ )}

sin (- x) + csc (x) = cot (x) cos (x)

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