Is csc(u)-sin(u)=cot(u)cos(u) ?

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

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prove csc(u)-sin(u)=cot(u)cos(u)

prove

csc (u) - sin (u) = cot (u) cos (u)

True

Solution steps

csc (u) - sin (u) = cot (u) cos (u)

Manipulating left side

csc (u) - sin (u)

Express with sin, cos

csc (u) - sin (u)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (u) sin (u)

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

sin (u) sin (u)

Apply exponent rule:

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

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

= sin^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= sin^{2} (u)

= 1 - sin^{2} (u)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (u) cot (u)

cos (u) cot (u)

= cot (u) cos (u)

We showed that the two sides could take the same form

\Rightarrow True

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

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

- \frac{1}{2} = sin (- \frac{π}{12}) 2 + 3

- sin (\frac{π}{4}) = - cos (\frac{π}{4})

tan (x) + cot (x) = tan (x)

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