Is (cot(x)+cos(x))/(1+csc(x))=cos(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (cot(x)+cos(x))/(1+csc(x))=cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Join

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

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

Convert element to fraction:

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

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

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

Cancel the common factor:

sin (x)

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

Factor out common term

cos (x)

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

Cancel the common factor:

sin (x) + 1

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (z + \frac{π}{2}) = cos (z)

\frac{sec ( α )}{1 - cos ( α )} = \frac{sec ( α ) + 1}{sin ^{2} ( α )}

2 csc (2 x) = csc (x) sec (x)

\frac{1}{sin ( y ) - 1} - \frac{1}{sin ( y ) + 1} = - 2 sec^{2} (y)

cos (\frac{π}{4}) = \frac{2}{2}

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