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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (1+cos(θ))/(sin(θ))=csc(θ)+cot(θ)

prove

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

True

Solution steps

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

Manipulating right side

csc (θ) + cot (θ)

Express with sin, cos

cot (θ) + csc (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

= \frac{cos ( θ ) + 1}{sin ( θ )}

= \frac{cos ( θ ) + 1}{sin ( θ )}

= \frac{1 + cos ( θ )}{sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

tan^{2} (θ) = sec^{2} (θ) - 1

(tan (x) + cot (x))^{2} = csc^{2} (x) sec^{2} (x)

(1 + sec (θ)) (1 - sec (θ)) = - tan^{2} (θ)

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

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