Is csc^2(x)-cos(x)sec(x)=cot^2(x) ?

The answer to whether csc^2(x)-cos(x)sec(x)=cot^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 csc^2(x)-cos(x)sec(x)=cot^2(x)

prove

csc^{2} (x) - cos (x) sec (x) = cot^{2} (x)

True

Solution steps

csc^{2} (x) - cos (x) sec (x) = cot^{2} (x)

Manipulating left side

csc^{2} (x) - cos (x) sec (x)

Rewrite using trig identities

csc^{2} (x) - cos (x) sec (x)

sec (x) cos (x) = 1

sec (x) cos (x)

Express with sin, cos

sec (x) cos (x)

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

= 1

= csc^{2} (x) - 1

Use the Pythagorean identity:

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

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

= cot^{2} (x)

= cot^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ^{2} ( a )}{1 - sin ^{2} ( a )} = sec^{2} (a) - 1

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

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

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

sin (2 θ) + 2 sin (θ) - cos (θ) = 2

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