Is 1/(csc(y)-cot(y))=csc(y)+cot(y) ?

The answer to whether 1/(csc(y)-cot(y))=csc(y)+cot(y) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove 1/(csc(y)-cot(y))=csc(y)+cot(y)

prove

\frac{1}{csc ( y ) - cot ( y )} = csc (y) + cot (y)

True

Solution steps

\frac{1}{csc ( y ) - cot ( y )} = csc (y) + cot (y)

Manipulating left side

\frac{1}{csc ( y ) - cot ( y )}

Rewrite using trig identities

\frac{1}{csc ( y ) - cot ( y )}

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

1 = csc^{2} (x) - cot^{2} (x)

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

Simplify

\frac{csc ^{2} ( y ) - cot ^{2} ( y )}{csc ( y ) - cot ( y )} : csc (y) + cot (y)

\frac{csc ^{2} ( y ) - cot ^{2} ( y )}{csc ( y ) - cot ( y )}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

csc^{2} (y) - cot^{2} (y) = (csc (y) + cot (y)) (csc (y) - cot (y))

= \frac{( csc ( y ) + cot ( y ) ) ( csc ( y ) - cot ( y ) )}{csc ( y ) - cot ( y )}

Cancel the common factor:

csc (y) - cot (y)

= csc (y) + cot (y)

= csc (y) + cot (y)

= csc (y) + cot (y)

We showed that the two sides could take the same form

\Rightarrow True

Is 1/(csc(y)-cot(y))=csc(y)+cot(y) ?
The answer to whether 1/(csc(y)-cot(y))=csc(y)+cot(y) is True