Is cos^2(θ)+tan^2(θ)cos^2(θ)=1 ?

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

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prove cos^2(θ)+tan^2(θ)cos^2(θ)=1

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (θ) + tan^{2} (θ) cos^{2} (θ)

Express with sin, cos

cos^{2} (θ) + cos^{2} (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ)

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

1 - cos (x) = sin (x)

csc (x) - sin (x) = cot (x) \cdot cos (x)

sin^{4} (y) - cos^{4} (y) = 1 - 2 cos^{2} (y)

1 - sin^{2} (- θ) = cos^{2} (θ)

tan (θ) csc^{2} (θ) - tan (θ) = cot (θ)

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