Question

Simplify the expression:

(cscy + coty)(cscy-coty) / cscy

Answer 1

Try using these trig identities to solve your problems:

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Answer 2

\(\bf \cfrac{[cos(y)+cot(y)][csc(y)-cot(y)]}{csc(y)} ?\)

Answer 3

[csc(y) + cot(y)][csc(y) - cot(y)] / csc(y)

Answer 4

hmmm actaully \(\bf \cfrac{[csc(y)+cot(y)][csc(y)-cot(y)]}{csc(y)} ?\) I meant
anyhow, had a typo

Answer 5

Yes

Answer 6

\(\bf \cfrac{[csc(y)+cot(y)][csc(y)-cot(y)]}{csc(y)}
\\ \quad \\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\ \quad \\ \quad \\
\cfrac{csc^2(y)-cot^2(y)}{csc(y)}\implies \cfrac{\cancel{csc^2(y)}}{\cancel{csc(y)}}- \cfrac{cot^2(y)}{csc(y)}
\\ \quad \\
csc(y)- \cfrac{cot^2(y)}{csc(y)}\implies \cfrac{1}{sin(y)}-\cfrac{\frac{cos^2(y)}{sin^2(y)}}{\frac{1}{sin(y)}}
\\ \quad \\
\cfrac{1}{sin(y)}-\cfrac{cos^2(y)}{\cancel{sin^2(y)}}\cdot \cfrac{\cancel{sin(y)}}{1}\implies \cfrac{1}{sin(y)}-\cfrac{cos^2(y)}{sin(y)}
\\ \quad \\
\cfrac{cos^2(y)}{sin^2(y)}\implies ?\)

Answer 7

Thank you!