Question

Prove:

Answer 1

\[\frac{ \sin^4x-\cos^4x }{ 1-\cot^4x }=\sin^4x\]

Answer 2

\[\sin^4 x-\cos^4 x=(\sin^2 x +\cos^2 x)(\sin^2 x - \cos^2 x) = \sin^2 x-\cos^2 x\]

Answer 3

\[1-\cot^4 x = (1-\cot^2 x)(1+\cot^2 x) = \csc^2 x (1-\cot^2 x)\]

Answer 4

my internet is working slow .....I cant see the equations u have written properly

Answer 5

can u draw?

Answer 6

Use the identity a^2-b^2 = (a-b)(a+b)

Answer 7

\[\frac{\sin^4x-\cos^4x}{1-\cot^4x}=\frac{\sin^4x-\cos^4x}{1-\frac{\cos^4x}{\sin^4x}}\\
=\frac{\sin^4x-\cos^4x}{\frac{\sin^4x-\cos^4x}{\sin^4x}}=\sin^4x
\]

Answer 8

It is rather pointless if one just give the answer right away.

Answer 9

I cannot see any of the equations listed above ^^

Answer 10

plz can someone draw....

Answer 11

the answer is right there.. just notice that the answer only has the sin^4x term. so, what is the factor on the other side that is causing the trouble? the cot..
so, convert it and viola

Answer 12

Nope, you see, we have \[\frac{\sin^2 x}{\csc^2 x}\]

Answer 13

:(

Answer 14

Here is a screenshot of the formulas typed in earlier:
(hope you can read them)

Answer 15

if you factorize it, you decrease the order of the system.
you final answer is still in sin^4
so, factorization will cause trouble.

Answer 16

@ZeHanz thanks it was kindof blind situation for me :D

Answer 17

@koli123able: It took a while for us to realize that :)

Answer 18

how did (sin^2x+cos^2x)(sin^2x-cos^2x) = sin^2x-cos^2x ??

Answer 19

@agostino ?

Answer 20

@ZeHanz the screenshot says --> (sin^2x+cos^2x)(sin^2x-cos^2x) = sin^2x-cos^2x

how did this happen??

Answer 21

It is because sin²x+cos²x=1 for all x.

Answer 22

what to do after sin^2x-cos^x/cosec^x(1-cot^2x)

Answer 23

@koli123able: although @electrokid received some critique about his giving away the answer, it is in fact the right one:

All this factoring and converting is a little over the top. Why?
Because what you have to prove is that the left hand side is equal to the rhs.
Everything there is about 4th powers. So there is no need to factor, and get lower than 4th powers.

Have a good look at his calculation and you will see what it is about.
To be able to see it, I'll include another screenshot :D

Answer 24

@koli123able and @agostino you are welcome
@ZeHanz Thanks

Answer 25

YW!