Is (csc^2(x)-1)/(csc(x))=cot(x)cos(x) ?

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

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prove (csc^2(x)-1)/(csc(x))=cot(x)cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{\frac{1}{s i n ( x )}}

Simplify

\frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{\frac{1}{s i n ( x )}} : \frac{- sin ^{2} ( x ) + 1}{sin ( x )}

\frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{\frac{1}{s i n ( x )}}

Apply the fraction rule:

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

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (x) cot (x)

cos (x) cot (x)

= cot (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (- x) + sin (- x) = cos (x) - sin (x)

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

cos (x - \frac{π}{3}) - sin (x + \frac{π}{6}) = 0

\frac{sin ( y )}{cot ( y )} = sec (y) - cos (y)

cos (x) - sin (x) = 2 \cdot cos (x + \frac{π}{4})

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