Is (1+cos(w))/(sin(w))=(sin(w))/(1-cos(w)) ?

The answer to whether (1+cos(w))/(sin(w))=(sin(w))/(1-cos(w)) is True

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prove (1+cos(w))/(sin(w))=(sin(w))/(1-cos(w))

prove

\frac{1 + cos ( w )}{sin ( w )} = \frac{sin ( w )}{1 - cos ( w )}

True

Solution steps

\frac{1 + cos ( w )}{sin ( w )} = \frac{sin ( w )}{1 - cos ( w )}

Manipulating right side

\frac{sin ( w )}{1 - cos ( w )}

Rewrite using trig identities

\frac{sin ( w )}{1 - cos ( w )}

Multiply by

\frac{1 + cos ( w )}{1 + cos ( w )}

= \frac{sin ( w )}{\frac{( 1 - c o s ( w ) ) ( 1 + c o s ( w ) )}{1 + c o s ( w )}}

Apply Difference of Two Squares Formula:

w^{2} - y^{2} = (w + y) (w - y)

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

= \frac{sin ( w )}{\frac{1 - c o s ^{2} ( w )}{1 + c o s ( w )}}

Use the Pythagorean identity:

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

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

= \frac{sin ( w )}{\frac{s i n ^{2} ( w )}{1 + c o s ( w )}}

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin (w)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{4 sec ( θ ) - 4}{1 - cos ( θ )} = 4 sec (θ)

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

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

tan (u) - 1 tan (u) + 1 = 1 - cot (u) 1 + cot (u)

cos (x) = sec (x) - sin (x) tan (x)

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