Is (sec(y))/(sec(y)-cos(y))=csc^2(y) ?

The answer to whether (sec(y))/(sec(y)-cos(y))=csc^2(y) is True

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prove (sec(y))/(sec(y)-cos(y))=csc^2(y)

Solution

prove

\frac{sec ( y )}{sec ( y ) - cos ( y )} = csc^{2} (y)

Solution

True

Solution steps

\frac{sec ( y )}{sec ( y ) - cos ( y )} = csc^{2} (y)

Manipulating left side

\frac{sec ( y )}{sec ( y ) - cos ( y )}

Express with sin, cos

\frac{sec ( y )}{- cos ( y ) + sec ( y )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{c o s ( y )}}{- cos ( y ) + \frac{1}{c o s ( y )}}

Simplify

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

\frac{\frac{1}{c o s ( y )}}{- cos ( y ) + \frac{1}{c o s ( y )}}

Apply the fraction rule:

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

= \frac{1}{cos ( y ) ( - cos ( y ) + \frac{1}{c o s ( y )} )}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (y) cos (y) + 1

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

cos (y) cos (y)

Apply exponent rule:

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

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

= cos^{1 + 1} (y)

Add the numbers:

1 + 1 = 2

= cos^{2} (y)

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (y)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{1}{sin ^{2} ( y )}

= \frac{1}{sin ^{2} ( y )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{( \frac{1}{c s c ( y )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c s c ( y )} ) ^{2}}

(\frac{1}{csc ( y )})^{2} = \frac{1}{csc ^{2} ( y )}

(\frac{1}{csc ( y )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( y )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( y )}

= \frac{1}{\frac{1}{c s c ^{2} ( y )}}

Apply the fraction rule:

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

= \frac{csc ^{2} ( y )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (y)

csc^{2} (y)

csc^{2} (y)

We showed that the two sides could take the same form

\Rightarrow True

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