Is tan^2(θ)cos^2(θ)=1-cos^2(θ) ?

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

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prove tan^2(θ)cos^2(θ)=1-cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (θ) cos^{2} (θ)

Express with sin, cos

cos^{2} (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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)

= sin^{2} (θ)

= sin^{2} (θ)

Rewrite using trig identities

sin^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x )}{sin ( 2 x )} = \frac{sec ( x )}{2}

sin (2 x) = 2 sin (x) \cdot cos (x)

tan (a) \frac{1}{sin ^{2} ( a )} = csc (a) sec (a)

(1 + csc (x)) (1 + csc (- x)) = - cot^{4} (x)

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

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