Question

verifying identities

(sin^2θ - tan θ) / (cos^2 θ - cot θ ) = tan ^2θ

Answer 1

put
\[\cos(x)=a, \sin(x)=b\] and then do some algebra. makes your life easier

Answer 2

\[\frac{b^2-\frac{b}{a}}{a^2-\frac{a}{b}}\] is what i mean
simplify this first before finding that it will be equal to the right hand side, which is
\[\frac{b^2}{a^2}\]

Answer 3

i would multiply top and bottom by ab first to clear the fractions
\[\frac{ab^3 -b^2}{a^3b-a^2}\]
\[\frac{b^2(ab-1)}{a^2(ab-1)}=\frac{b^2}{a^2}\] now replace "a" by cosine and "b" by sine and you get
\[\frac{\sin^2(x)}{\cos^2(x)}=\tan^2(x)\] so you see that there is not trigonometry in this problem, it is all algebra