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

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

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prove 1/(sin^2(x)tan(x))=csc^2(x)cot(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{sin ^{2} ( x ) \frac{s i n ( x )}{c o s ( x )}} : \frac{cos ( x )}{sin ^{3} ( x )}

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

Multiply

sin^{2} (x) \frac{sin ( x )}{cos ( x )} : \frac{sin ^{3} ( x )}{cos ( x )}

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

Multiply fractions:

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

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

sin (x) sin^{2} (x) = sin^{3} (x)

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

Apply exponent rule:

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

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

= sin^{1 + 2} (x)

Add the numbers:

1 + 2 = 3

= sin^{3} (x)

= \frac{sin ^{3} ( x )}{cos ( x )}

= \frac{1}{\frac{s i n ^{3} ( x )}{c o s ( x )}}

Apply the fraction rule:

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

= \frac{cos ( x )}{sin ^{3} ( x )}

= \frac{cos ( x )}{sin ^{3} ( x )}

= \frac{cos ( x )}{sin ^{3} ( x )}

Manipulating right side

csc^{2} (x) cot (x)

Express with sin, cos

cot (x) csc^{2} (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{cos ( x )}{sin ( x )} (\frac{1}{sin ( x )})^{2} : \frac{cos ( x )}{sin ^{3} ( x )}

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

Multiply fractions:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

Apply the fraction rule:

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

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

sin^{2} (x) sin (x) = sin^{3} (x)

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

Apply exponent rule:

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

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

= sin^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= sin^{3} (x)

= \frac{cos ( x )}{sin ^{3} ( x )}

= \frac{cos ( x )}{sin ^{3} ( x )}

= \frac{cos ( x )}{sin ^{3} ( x )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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