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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{sin ( x ) - csc ( x )}{csc ( x )}

Express with sin, cos

\frac{- csc ( x ) + sin ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

= \frac{- \frac{1}{s i n ( x )} + sin ( x )}{\frac{1}{s i n ( x )}}

Simplify

\frac{- \frac{1}{s i n ( x )} + sin ( x )}{\frac{1}{s i n ( x )}} : - 1 + sin^{2} (x)

\frac{- \frac{1}{s i n ( x )} + sin ( x )}{\frac{1}{s i n ( x )}}

Apply the fraction rule:

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

= \frac{( - \frac{1}{s i n ( x )} + sin ( x ) ) sin ( x )}{1}

Join

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

- \frac{1}{sin ( x )} + sin (x)

Convert element to fraction:

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

= - \frac{1}{sin ( x )} + \frac{sin ( x ) sin ( x )}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 + sin ( x ) sin ( x )}{sin ( x )}

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

- 1 + sin (x) sin (x)

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

= \frac{\frac{s i n ^{2} ( x ) - 1}{s i n ( x )} sin ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

Rewrite using trig identities

- 1 + sin^{2} (x)

Use the Pythagorean identity:

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

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

= - cos^{2} (x)

= - cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

cot^{2} (θ) csc^{2} (θ) - cot^{2} (θ) = cot^{4} (θ)

sin (x - π) = - sin (x)

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

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