Question

(1-cos^2x)(1-tan^2x) = sin^x-2sin^4x/1-sin^2x prove ls=rs

Answer 1

I think there is a typo in the rhs
what is
sin^x
?

Answer 2

sin^2x

Answer 3

\[(1-\cos^2x)(1-\tan^2x)={\sin^2x-2\sin^4x\over1-\sin^2x}\]right?

Answer 4

Yes

Answer 5

I did left side, and so far I got sin^2x (1-sin^2x/cos^2x) is taht right?

Answer 6

yeah that is correct, though I think it may be easier to work the right side first...

Answer 7

I trued but didn't really know what to do, I got stuck

Answer 8

*tried

Answer 9

that 2 doesn't seem to belong...
are you sure there is a coefficient of 2 on the right?

Answer 10

yess

Answer 11

\[(1-\cos^2x)(1-\tan^2x)=\sin^2x(1-{\sin^2x\over\cos^2x})=\sin^2x-{\sin^4x\over\cos^2x}\]oh yes, that is right
now take cos^2x as a common denominator so you can put the fractions together, then use cos^2x=1-sin^2x

Answer 12

\[(1-\cos^2x)(1-\tan^2x)=\sin^2x(1-{\sin^2x\over\cos^2x})=\sin^2x-{\sin^4x\over\cos^2x}\]\[={\sin^2x\cos^2x-\sin^4x\over\cos^2x}\]

Answer 13

now like I say\[\cos^2x=1-\sin^2x\]and you're done

Answer 14

wait, so you just made the top cos^2x and bottom cos^2x into 1-sin^2x and brought it to the bottom?

Answer 15

and but it answer should come up to sin^2x-2sin^2x/1-sin^2x

Answer 16

\[(1-\cos^2x)(1-\tan^2x)=\sin^2x(1-{\sin^2x\over\cos^2x})=\sin^2x-{\sin^4x\over\cos^2x}\]\[=\sin^2x\cdot{\cos^2x\over\cos^2x}-{\sin^4x\over\cos^2x}={\sin^2x\cos^2x-\sin^4x\over\cos^2x}\]now \[\cos^2x=1-\sin^2x\]and simplify and you do get that answer

Answer 17

I multiplied the \(\sin^2x\) by\[\frac{\cos^2x}{\cos^2x}\]

Answer 18

oh okayy, thank youu!

Answer 19

welcome :)