Is sec^2(x)-sin^2(x)sec^2(x)=1 ?

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

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prove sec^2(x)-sin^2(x)sec^2(x)=1

prove

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

True

Solution steps

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

Manipulating left side

sec^{2} (x) - sin^{2} (x) sec^{2} (x)

Express with sin, cos

sec^{2} (x) - sec^{2} (x) sin^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

tan (2 a) - 2 tan (2 a) sin^{2} (a) = sin (2 a)

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

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

80 = arcsin (45.6443 \cdot \frac{sin ( c )}{35})

sin (0) csc (0) = 1

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