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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{sin ( x )}

Simplify

\frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{sin ( x )} : \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x ) sin ( x )}

\frac{- 1 + ( \frac{1}{c o s ( x )} ) ^{2}}{sin ( 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 + \frac{1}{c o s ^{2} ( x )}}{sin ( x )}

Join

- 1 + \frac{1}{cos ^{2} ( x )} : \frac{- cos ^{2} ( x ) + 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{\frac{- c o s ^{2} ( x ) + 1}{c o s ^{2} ( x )}}{sin ( x )}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

1 - (1 - sin^{2} (x)) : sin^{2} (x)

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= sin^{2} (x)

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

Cancel the common factor:

sin (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (\frac{π}{4} + x) + sin (\frac{π}{4} - x) = 2 cos (x)

p ro v e sin (θ) sec (θ) cot (θ) = 1

p ro v e 1 + sec^{2} (θ) sin^{2} (θ) = sec^{2} (θ)

p ro v e 3 sin^{2} (x) + 4 cos^{2} (x) = 3 + cos^{2} (x)

p ro v e tan (\frac{π}{2} - θ) tan (θ) = 1

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