Is 1/(cot^2(x)sin^2(x))=1+tan^2(x) ?

The answer to whether 1/(cot^2(x)sin^2(x))=1+tan^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 1/(cot^2(x)sin^2(x))=1+tan^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1}{( \frac{c o s ( x )}{s i n ( x )} ) ^{2} sin ^{2} ( x )}

Simplify

\frac{1}{( \frac{c o s ( x )}{s i n ( x )} ) ^{2} sin ^{2} ( x )} : \frac{1}{cos ^{2} ( x )}

\frac{1}{( \frac{c o s ( x )}{s i n ( x )} ) ^{2} sin ^{2} ( x )}

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

= \frac{1}{\frac{c o s ^{2} ( x )}{s i n ^{2} ( x )} sin ^{2} ( x )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

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

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

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

Manipulating right side

1 + tan^{2} (x)

Express with sin, cos

1 + tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sec^{4} (x) - tan^{4} (x) = 2 tan^{2} (x) + 1

1 - 2 cos^{2} (θ) + cos^{4} (θ) = sin^{4} (θ)

csc^{2} (θ) sec^{2} (θ) - csc^{2} (θ) = sec^{2} (θ)

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

tan (a) + cot (a) = \frac{csc ( a )}{cos ( a )}

Is 1/(cot^2(x)sin^2(x))=1+tan^2(x) ?
The answer to whether 1/(cot^2(x)sin^2(x))=1+tan^2(x) is True