Is 1-tan^2(A)cos^2(A)=cos^2(A) ?

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

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prove 1-tan^2(A)cos^2(A)=cos^2(A)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (A)

= sin^{2} (A)

= 1 - sin^{2} (A)

= 1 - sin^{2} (A)

= 1 - sin^{2} (A)

Rewrite using trig identities

1 - sin^{2} (A)

Use the Pythagorean identity:

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

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

= cos^{2} (A)

= cos^{2} (A)

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{1 - 4 sec ^{2} ( x )}{1 + 2 sec ( x )} = 1 - 2 sec (x)

sin (\frac{3 π}{4}) = \frac{2}{2}

tan (x) = cot (\frac{π}{2} - x)

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