Is sec^2(θ)-sin^2(θ)sec^2(θ)=1 ?

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

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prove sec^2(θ)-sin^2(θ)sec^2(θ)=1

prove

sec^{2} (θ) - sin^{2} (θ) sec^{2} (θ) = 1

True

Solution steps

sec^{2} (θ) - sin^{2} (θ) sec^{2} (θ) = 1

Manipulating left side

sec^{2} (θ) - sin^{2} (θ) sec^{2} (θ)

Express with sin, cos

sec^{2} (θ) - sec^{2} (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{cos ( θ )})^{2} sin^{2} (θ)

(\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 )} sin^{2} (θ)

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

9 cot^{2} (y) (sec^{2} (y) - 1) = 9

sin (x) cos (x) = \frac{1}{2} sin (2 x)

\frac{tan ^{2} ( t ) - 1}{sec ^{2} ( t )} = \frac{tan ( t ) - cot ( t )}{tan ( t ) + cot ( t )}

8 sin (x) cos (x) = 4 sin (2 x)

tan (x) cos (x) - \frac{1}{csc ( x )} = 0

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