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

The answer to whether csc(x)sec(x)cot(x)=csc^2(x) is True

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prove csc(x)sec(x)cot(x)=csc^2(x)

prove

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

True

Solution steps

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

Manipulating left side

csc (x) sec (x) cot (x)

Express with sin, cos

cot (x) csc (x) sec (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{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

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

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

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

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

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

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