Is cos^2(x)+tan^2(x)cos^2(x)=1 ?

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

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prove cos^2(x)+tan^2(x)cos^2(x)=1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

cos^{2} (x) + 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)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

cos (x - \frac{π}{3}) + sin (\frac{π}{6} - x) = cos (x)

coth^{2} (x) - 1 = csch^{2} (x)

\frac{cos ^{4} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )} = 1 - tan^{2} (x)

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

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