Is tan(x)cot(x)-cos^2(x)=sin^2(x) ?

The answer to whether tan(x)cot(x)-cos^2(x)=sin^2(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan(x)cot(x)-cos^2(x)=sin^2(x)

prove

tan (x) cot (x) - cos^{2} (x) = sin^{2} (x)

True

Solution steps

tan (x) cot (x) - cos^{2} (x) = sin^{2} (x)

Manipulating left side

tan (x) cot (x) - cos^{2} (x)

Express with sin, cos

- cos^{2} (x) + cot (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

Cancel the common factor:

sin (x)

= 1

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

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

= 1 - cos^{2} (x)

Rewrite using trig identities

1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (t) - cos^{2} (t) = cot^{2} (t) cos^{2} (t)

cot (θ) = \frac{cos ( θ )}{sin ( θ )}

(1 - cot (x))^{2} = csc^{2} (x) - 2 cot (x)

\frac{1}{cot ( x ) ( 1 - cos ( 2 x ) )} = csc (2 x)

tan^{4} (x) = tan^{2} (x) sec^{2} (x) - tan^{2} (x)

Is tan(x)cot(x)-cos^2(x)=sin^2(x) ?
The answer to whether tan(x)cot(x)-cos^2(x)=sin^2(x) is True