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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

prove

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

True

Solution steps

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

Manipulating right side

csc (v) - sin (v)

Express with sin, cos

csc (v) - sin (v)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (v) sin (v)

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

sin (v) sin (v)

Apply exponent rule:

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

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

= sin^{1 + 1} (v)

Add the numbers:

1 + 1 = 2

= sin^{2} (v)

= 1 - sin^{2} (v)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (2 x) = 2 cot (x) sin^{2} (x)

p ro v e \frac{sec ( t ) - cos ( t )}{sec ( t )} = sin^{2} (t)

p ro v e \frac{1}{tan ( x )} + tan (x) = sec (x) csc (x)

p ro v e tan (θ) + cot (θ) = 2 csc (2 θ)

p ro v e csc^{4} (x) - cot^{4} (x) = 2 cot^{2} (x) + 1

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