Is (cos^2(t))/(sin(t))=csc(t)-sin(t) ?

The answer to whether (cos^2(t))/(sin(t))=csc(t)-sin(t) is True

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prove (cos^2(t))/(sin(t))=csc(t)-sin(t)

prove

\frac{cos ^{2} ( t )}{sin ( t )} = csc (t) - sin (t)

True

Solution steps

\frac{cos ^{2} ( t )}{sin ( t )} = csc (t) - sin (t)

Manipulating right side

csc (t) - sin (t)

Express with sin, cos

csc (t) - sin (t)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (t) sin (t)

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

sin (t) sin (t)

Apply exponent rule:

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

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

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

= 1 - sin^{2} (t)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos (x - \frac{π}{6}) - cos (x + \frac{π}{6}) = sin (x)

cos (\frac{3 π}{2} + θ) = sin (θ)

csc^{2} (θ) - cos^{2} (θ) csc^{2} (θ) = 1

Is (cos^2(t))/(sin(t))=csc(t)-sin(t) ?
The answer to whether (cos^2(t))/(sin(t))=csc(t)-sin(t) is True