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

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

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

prove

\frac{1 + tan ( x )}{1 + \frac{1}{t a n ( x )}} = tan (x)

True

Solution steps

\frac{1 + tan ( x )}{1 + \frac{1}{t a n ( x )}} = tan (x)

Manipulating left side

\frac{1 + tan ( x )}{1 + \frac{1}{t a n ( x )}}

Simplify

\frac{1 + tan ( x )}{1 + \frac{1}{t a n ( x )}} : tan (x)

\frac{1 + tan ( x )}{1 + \frac{1}{t a n ( x )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot tan (x) = tan (x)

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

= \frac{1 + tan ( x )}{\frac{t a n ( x ) + 1}{t a n ( x )}}

Apply the fraction rule:

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

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

Cancel the common factor:

1 + tan (x)

= tan (x)

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (2 θ) = \frac{1}{sec ( 2 θ )}

\frac{1}{csc ^{2} ( A )} + \frac{1}{cot ^{2} ( A )} = 1

(cos (3 x) - cos (x)) = - 2 sin (2 x) sin (x)

sec (\frac{π}{2} - y) = csc (y)

cos (2 x - \frac{π}{2}) = cos (\frac{π}{2} - 2 x)

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