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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin (x) - cos (x) = 1 - sin (2 x)

9 sec^{2} (x) - 5 tan^{2} (x) = 5 + 4 sec^{2} (x)

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

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