Is csc(ω)-sin(ω)=cot(ω)cos(ω) ?

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

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prove csc(ω)-sin(ω)=cot(ω)cos(ω)

prove

csc (ω) - sin (ω) = cot (ω) cos (ω)

True

Solution steps

csc (ω) - sin (ω) = cot (ω) cos (ω)

Manipulating left side

csc (ω) - sin (ω)

Express with sin, cos

csc (ω) - sin (ω)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( ω )} - sin (ω)

Simplify

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

\frac{1}{sin ( ω )} - sin (ω)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - sin (ω) sin (ω) = 1 - sin^{2} (ω)

1 - sin (ω) sin (ω)

sin (ω) sin (ω) = sin^{2} (ω)

sin (ω) sin (ω)

Apply exponent rule:

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

sin (ω) sin (ω) = sin^{1 + 1} (ω)

= sin^{1 + 1} (ω)

Add the numbers:

1 + 1 = 2

= sin^{2} (ω)

= 1 - sin^{2} (ω)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( ω )}{sin ( ω )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( ω )}{sin ( ω )}

= \frac{cos ^{2} ( ω )}{sin ( ω )}

Rewrite using trig identities

= cos (ω) \frac{cos ( ω )}{sin ( ω )}

Use the basic trigonometric identity:

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

cos (ω) cot (ω)

cos (ω) cot (ω)

= cot (ω) cos (ω)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ( 3 θ ) - tan ( 3 θ )}{sin ( 3 θ ) + cos ( 3 θ )} = csc (3 θ) - sec (3 θ)

sin (- \frac{π}{2} - θ) = - cos (θ)

- 2 cos (2 x) \cdot sin (2 x) = sin (- 4 x)

sec^{2} (x) + tan (x) - 1 = 0

tan (a) cot (a) = 1

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