Is csc(x)-cos(x)*cot(x)=sin(x) ?

The answer to whether csc(x)-cos(x)*cot(x)=sin(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove csc(x)-cos(x)*cot(x)=sin(x)

prove

True

Solution steps

csc (x) - cos (x) cot (x) = sin (x)

Manipulating left side

csc (x) - cos (x) cot (x)

Express with sin, cos

csc (x) - cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{sin ( x )} - cos (x) \frac{cos ( x )}{sin ( x )} : \frac{1 - cos ^{2} ( x )}{sin ( x )}

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

cos (x) \frac{cos ( x )}{sin ( x )} = \frac{cos ^{2} ( x )}{sin ( x )}

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

Multiply fractions:

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

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

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= \frac{cos ^{2} ( x )}{sin ( x )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( x )}{sin ( x )}

Use the Pythagorean identity:

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

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

= \frac{sin ^{2} ( x )}{sin ( x )}

Cancel the common factor:

sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

Is csc(x)-cos(x)*cot(x)=sin(x) ?
The answer to whether csc(x)-cos(x)*cot(x)=sin(x) is True