Is csc^2(x)=(sec^2(x))/(tan^2(x)) ?

The answer to whether csc^2(x)=(sec^2(x))/(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 csc^2(x)=(sec^2(x))/(tan^2(x))

prove

csc^{2} (x) = \frac{sec ^{2} ( x )}{tan ^{2} ( x )}

True

Solution steps

csc^{2} (x) = \frac{sec ^{2} ( x )}{tan ^{2} ( x )}

Manipulating right side

\frac{sec ^{2} ( x )}{tan ^{2} ( x )}

Express with sin, cos

\frac{sec ^{2} ( x )}{tan ^{2} ( x )}

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( x )} ) ^{2}}{tan ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{( \frac{1}{c s c ( x )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c s c ( x )} ) ^{2}}

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

(\frac{1}{csc ( x )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{c s c ^{2} ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= csc^{2} (x)

csc^{2} (x)

csc^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

1 = sec^{2} (θ) - tan^{2} (θ)

\frac{1}{sin ( θ ) cot ( θ )} = \frac{1}{cos ( θ )}

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

2 (sin (2 y) cos (2 y)) = sin (4 y)

Is csc^2(x)=(sec^2(x))/(tan^2(x)) ?
The answer to whether csc^2(x)=(sec^2(x))/(tan^2(x)) is True