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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec^{2} (x) - 1

Express with sin, cos

- 1 + sec^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

tan (a) + cot (a) = 2 csc (2 a)

\frac{sec ^{2} ( t )}{csc ( t )} = sec (t) tan (t)

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

3 cos^{2} (x) - 3 sin^{2} (x) = 3 - 6 sin^{2} (x)

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

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