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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

= - \frac{cos ( x )}{sin ( x )} + \frac{1 + \frac{1}{s i n ( x )}}{sec ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

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

Apply rule

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

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

Factor

- cos (x) + (sin (x) + 1) cos (x) : cos (x) sin (x)

- cos (x) + (sin (x) + 1) cos (x)

Factor out common term

cos (x)

= cos (x) (- 1 + 1 + sin (x))

Refine

= cos (x) sin (x)

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

Cancel the common factor:

sin (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

p ro v e \frac{csc ( - x )}{sec ( - x )} = - cot (x)

p ro v e cos (x - y) + cos (x + y) = 2 cos (x) cos (y)

p ro v e sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos (x)

p ro v e (cot (x) - csc (x)) (cos (x) + 1) = - sin (x)

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