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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (x) cos (x) + 1

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Cancel the common factor:

sin^{2} (x)

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( x )}}

Simplify

\frac{1}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin^{3} (θ) + sin (θ) cos^{2} (θ) = sin (θ)

sec^{2} (θ) cot^{2} (θ) = csc^{2} (θ)

tan (\frac{3 π}{2} - x) = cot (x)

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