Is sec^2(θ)(1-cos^2(θ))=tan^2(θ) ?

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

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prove sec^2(θ)(1-cos^2(θ))=tan^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sec^{2} (θ) (1 - cos^{2} (θ))

Express with sin, cos

(1 - cos^{2} (θ)) sec^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

(1 - cos^{2} (θ)) (\frac{1}{cos ( θ )})^{2}

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Multiply fractions:

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

= \frac{1 \cdot ( 1 - cos ^{2} ( θ ) )}{cos ^{2} ( θ )}

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

1 \cdot (1 - cos^{2} (θ))

Multiply:

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

= (1 - cos^{2} (θ))

Remove parentheses:

(a) = a

= 1 - cos^{2} (θ)

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

= \frac{sin ( θ )}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

\frac{sin ( θ ) tan ( θ )}{cos ( θ )}

= tan (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

tan (θ) tan (θ)

Simplify

tan (θ) tan (θ) : tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

tan^{2} (θ)

tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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)

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