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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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)

= sin^{2} (x)

= sin^{2} (x)

Rewrite using trig identities

sin^{2} (x)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (A) - sin^{2} (A) cos^{2} (A) = sin^{4} (A)

\frac{1}{sin ( x ) cos ( x )} = sec (x) csc (x)

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

sin (a) sec (a) tan (a) + 1 = sec^{2} (a)

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

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