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

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

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

prove

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

True

Solution steps

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

Manipulating right side

1 + tan (x)

Express with sin, cos

1 + tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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{cos ( x ) + sin ( x )}{cos ( x )}

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{1 + sin ( - x )}{cos ( x ) tan ( x ) - 1} = - 1

p ro v e sec (A) - \frac{cos ( A )}{1 + sin ( A )} = tan (A)

p ro v e \frac{1 - tan ^{2} ( A )}{1 + tan ^{2} ( A )} = cos (2 A)

p ro v e sin (\frac{π}{3}) = cos (\frac{π}{6})

p ro v e cos^{2} (t) = \frac{( 1 + cos ( 2 t ) )}{2}

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