Is (1-cos^2(θ))/(tan^2(θ))=cos^2(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin^{2} (θ)

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (θ) cos^{2} (θ) = sin^{2} (θ)

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

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

\frac{tan ( a )}{sec ( a )} = sin (a)

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

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