Is (csc(x)-sin(x))/(cot(x))=cos(x) ?

The answer to whether (csc(x)-sin(x))/(cot(x))=cos(x) is True

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prove (csc(x)-sin(x))/(cot(x))=cos(x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ( x ) - sin ( x )}{cot ( x )}

Express with sin, cos

\frac{csc ( x ) - sin ( x )}{cot ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( x )} - sin ( x )}{cot ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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)

= 1 - sin^{2} (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= 1 - sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

cos^{2} (x) cot^{2} (x) = cot^{2} (x) - cos^{2} (x)

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

tan (4 5^{\circ} + x) + tan (4 5^{\circ} - x) = 2 sec (2 x)

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

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