Question

tan^2x-cot^2x=??
ill draw a pic

Answer 1

\(\LARGE\color{black}{ \frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x} }\)

Answer 2

\(\LARGE\color{black}{ \frac{\sin^2x\sin^2x}{\cos^2x\sin^2x}-\frac{\cos^2x\cos^2x}{\sin^2x\cos^2x} }\)

Answer 3

okay so common denominators and i just cancel what needs to be canceled?

Answer 4

you are right about the common denominator, but don't be too quick... tell me first, what do you get after adding the fractions.

Answer 5

\(\LARGE\color{black}{ \frac{\sin^2x\sin^2x}{\cos^2x\sin^2x}-\frac{\cos^2x\cos^2x}{\sin^2x\cos^2x} }\)

\(\LARGE\color{black}{ \frac{\sin^2x\sin^2x-\cos^2x\cos^2x}{\cos^2x\sin^2x}}\)

Answer 6

\(\LARGE\color{black}{ \frac{\sin^2x(1-\cos^2x)-\cos^2x(1-\sin^2x)}{\cos^2x\sin^2x}}\)

\(\LARGE\color{black}{ \frac{\sin^2x-\sin^2x\cos^2x~~-~~\cos^2x-\sin^2x\cos^2x}{\cos^2x\sin^2x}}\)

Answer 7

\(\LARGE\color{black}{ \frac{\sin^2x-2\sin^2x\cos^2x-\cos^2x}{\cos^2x\sin^2x}}\)

Answer 8

\(\LARGE\color{black}{ \frac{\sin^2x-2\sin^2x\cos^2x-(1-\sin^2x)}{\cos^2x\sin^2x}}\)

\(\LARGE\color{black}{ \frac{\sin^2x-2\sin^2x\cos^2x-1+\sin^2x}{\cos^2x\sin^2x}}\)

Answer 9

\(\LARGE\color{black}{ \frac{2\sin^2x-2\sin^2x\cos^2x-1}{\cos^2x\sin^2x}}\)

Answer 10

then just re-write each term.

Answer 11

okay got it thanks :)

Answer 12

just in case,