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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply the fraction rule:

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

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

Join

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

1 + \frac{1}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ ) + 1}{sin ( θ )}

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

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

Apply rule

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

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

Factor

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

- cos (θ) + (sin (θ) + 1) cos (θ)

Factor out common term

cos (θ)

= cos (θ) (- 1 + 1 + sin (θ))

Refine

= cos (θ) sin (θ)

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

Cancel the common factor:

sin (θ)

= cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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