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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan (u) (csc (u) - sin (u))

Express with sin, cos

(csc (u) - sin (u)) tan (u)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Join

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (u)

= 1 - sin^{2} (u)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (u)

= cos (u)

= cos (u)

We showed that the two sides could take the same form

\Rightarrow True

sin (x - y) + sin (x + y) = 2 cos (y) sin (x)

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

1 - 2 sin^{2} (x) + sin^{4} (x) = cos^{4} (x)

\frac{3 sin ^{2} ( t )}{tan ^{2} ( t )} = 3 cos^{2} (t)

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

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