Is (sec(x)+1)/(sec(x))=1+cos(x) ?

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

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prove (sec(x)+1)/(sec(x))=1+cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= cos (x) + 1

= cos (x) + 1

= 1 + cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

- 3 cos^{2} (θ) + 11 = - 5 sin (θ) + 6

(1 - tan (x))^{2} = 1 - 2 tan (x) + tan^{2} (x)

csc (x) cos (x) tan (x) = 1

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

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