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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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 )}{sin ( x )}

Multiply

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

\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

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

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

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

Manipulating right side

cot (x) + csc (x)

Express with sin, cos

cot (x) + csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

Apply rule

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

cos (π + x) = cos (x)

sec (2 x) = \frac{( sec ^{2} ( x ) )}{1 - tan ^{2} ( x )}

(sin^{2} (x) - 1)^{2} = cos (2 x) + sin^{4} (x)

sec (x) - cos (x) = sin^{2} (x) sec (x)

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