Is (sin(θ))/(cot(θ)+csc(θ))=1-cos(θ) ?

The answer to whether (sin(θ))/(cot(θ)+csc(θ))=1-cos(θ) is True

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prove (sin(θ))/(cot(θ)+csc(θ))=1-cos(θ)

prove

\frac{sin ( θ )}{cot ( θ ) + csc ( θ )} = 1 - cos (θ)

True

Solution steps

\frac{sin ( θ )}{cot ( θ ) + csc ( θ )} = 1 - cos (θ)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{sin ( θ )}{\frac{c o s ( θ )}{s i n ( θ )} + csc ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

\frac{cos ( θ )}{sin ( θ )} + \frac{1}{sin ( θ )} : \frac{cos ( θ ) + 1}{sin ( θ )}

Apply rule

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

= \frac{cos ( θ ) + 1}{sin ( θ )}

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

Apply the fraction rule:

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

= \frac{sin ( θ ) sin ( θ )}{cos ( θ ) + 1}

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)

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

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

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

Manipulating right side

1 - cos (θ)

Rewrite using trig identities

1 - cos (θ)

Multiply by

\frac{1 + cos ( θ )}{1 + cos ( θ )}

= \frac{( 1 - cos ( θ ) ) ( 1 + cos ( θ ) )}{1 + cos ( θ )}

Apply Difference of Two Squares Formula:

θ^{2} - y^{2} = (θ + y) (θ - y)

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

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

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

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

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (x) cos^{2} (x) = cot^{2} (x)

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

1 + tan^{2} (θ) = \frac{1}{cos ^{2} ( θ )}

cos^{2} (B) = \frac{1}{2} (1 + cos (2 B))

sin (θ + π) = - sin (θ)

Is (sin(θ))/(cot(θ)+csc(θ))=1-cos(θ) ?
The answer to whether (sin(θ))/(cot(θ)+csc(θ))=1-cos(θ) is True