Is (csc(θ)-sin(θ))/(cot(θ))=cos(θ) ?

The answer to whether (csc(θ)-sin(θ))/(cot(θ))=cos(θ) is True

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prove (csc(θ)-sin(θ))/(cot(θ))=cos(θ)

prove

\frac{csc ( θ ) - sin ( θ )}{cot ( θ )} = cos (θ)

True

Solution steps

\frac{csc ( θ ) - sin ( θ )}{cot ( θ )} = cos (θ)

Manipulating left side

\frac{csc ( θ ) - sin ( θ )}{cot ( θ )}

Express with sin, cos

\frac{csc ( θ ) - sin ( θ )}{cot ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( θ )} - sin ( θ )}{cot ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( θ )} - sin ( θ )}{\frac{c o s ( θ )}{s i n ( θ )}}

Simplify

\frac{\frac{1}{s i n ( θ )} - sin ( θ )}{\frac{c o s ( θ )}{s i n ( θ )}} : \frac{1 - sin ^{2} ( θ )}{cos ( θ )}

\frac{\frac{1}{s i n ( θ )} - sin ( θ )}{\frac{c o s ( θ )}{s i n ( θ )}}

Apply the fraction rule:

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

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

Join

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

\frac{1}{sin ( θ )} - sin (θ)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - sin (θ) sin (θ) = 1 - sin^{2} (θ)

1 - sin (θ) sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= 1 - sin^{2} (θ)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= 1 - sin^{2} (θ)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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

cot (x) = arctan (x)

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

sin (x) tan (x) = 1 + cos (x)

Is (csc(θ)-sin(θ))/(cot(θ))=cos(θ) ?
The answer to whether (csc(θ)-sin(θ))/(cot(θ))=cos(θ) is True