Question

prove: sec^2x-1/secx-1 = tan^2x/secx-1

Answer 1

\[\frac{ \sec ^{2}x-1 }{ \sec x-1 }\]The numerator is actually a trig identity of
\[\cos ^{2}x+\sin ^{2}x=1\]When you divide everything by cos^2x you get
\[1+\tan ^{2}x=\sec ^{2}x\]Thus when you minus the one you get
\[\tan ^{2}x=\sec ^{2}x-1\]Then that's how they got
\[\frac{ \tan ^{2}x }{ \sec x-1 }\]
Hope it helps and Good Luck :)

Answer 2

Thank you!

Answer 3

No Problem