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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{cot ( x )}{sec ( x )}

Express with sin, cos

\frac{cot ( x )}{sec ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )}}{sec ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Refine

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Manipulating right side

csc (x) - sin (x)

Express with sin, cos

csc (x) - sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (x) (sec^{2} (x) - 1) = 1

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

sec (x) + tan (x) = \frac{1}{sec ( x ) - tan ( x )}

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

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

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