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

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

= sin^{2} (x)

= 1 - sin^{2} (x)

= 1 - sin^{2} (x)

= 1 - sin^{2} (x)

Rewrite using trig identities

1 - sin^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (2 a) = \frac{1}{2} (cot (a) - tan (a))

sec (t) = \frac{cot ( t ) + tan ( t )}{csc ( t )}

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

sin (2 x) + sin (x) = sin (3 x)

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