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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ( x ) - sin ( 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

sin (a) = \frac{2 tan ( \frac{a}{2} )}{1 + tan ^{2} ( \frac{a}{2} )}

sin (θ) = sin (θ - 2 π)

3 sec^{2} (x) - 2 tan^{2} (x) - 4 = 0

tan (θ) cot (θ) - 1 = 1 - sec (θ) cos (θ)

cos (2 θ) + tan (θ) sin (2 θ) = 1

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