Is (1-cos^2(x))/(tan(x))=sin(x)cos(x) ?

The answer to whether (1-cos^2(x))/(tan(x))=sin(x)cos(x) is True

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prove (1-cos^2(x))/(tan(x))=sin(x)cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1 - cos ^{2} ( x )}{\frac{s i n ( x )}{c o s ( x )}}

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

= sin (x) cos (x)

= sin (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

cot (θ) \cdot sec (θ) = csc (θ)

csc^{2} (x) + cot^{2} (x) = 1 + 2 cot^{2} (x)

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

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

Is (1-cos^2(x))/(tan(x))=sin(x)cos(x) ?
The answer to whether (1-cos^2(x))/(tan(x))=sin(x)cos(x) is True