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

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

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

prove

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

True

Solution steps

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

Manipulating right 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 )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{sin ^{2} ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

Simplify

\frac{sin ^{2} ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{sin ^{2} ( x )}{\frac{1}{s e c ^{2} ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

(6 cos^{2} (x) - 6 sin^{2} (x)) = 6 - 12 sin^{2} (x)

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

cos^{2} (x) - 4 = 4 - sin^{2} (x)

sec (x \cdot 0) = 1

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

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