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

The answer to whether sin(x)-csc(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

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

sin (x) - csc (x)

Express with sin, cos

- csc (x) + sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- 1 + sin (x) sin (x)

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

- cos (x) cot (x)

- cos (x) cot (x)

= - cot (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

1 - sin (θ) cos (θ) tan (θ) = cos^{2} (θ)

tan (x) \cdot cos (x) \cdot csc (x) = 1

\frac{1}{cot ^{2} ( x )} = \frac{tan ( x )}{cot ( x )}

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

\frac{cot ^{2} ( x )}{1 + cot ^{2} ( x )} = cos^{2} (x)

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