Question

prove that this is an identity... sin^2x(cotx+1)^2=cos^2x(tanx+1)^2
Please help, these are really confusing.

Answer 1

thanks!

Answer 2

I have it too

Answer 3

nvm

Answer 4

sorry

Answer 5

@hero does it usually take this long to get the solution ? I got a bunch of these to do =/

Answer 6

There's a specific approach to this that works most simply.

First apply \(a^2b^2 = (ab)^2\)

\(\sin^2x(\cot(x) + 1)^2 = [\sin(x)(\cot(x) + 1]^2\)

Answer 7

No, it shouldn't take too long if you are very familiar with these. I already explained to you that my instructor allowed me to do these a different way so I had a different experience than most students.

Answer 8

Okay so now we have

\[[\sin(x)(\cot(x) + 1)]^2\]

Change \(\cot(x)\) to \(\dfrac{\cos(x)}{\sin(x)}\)

Answer 9

\[\left[\sin(x)\left(\frac{\cos(x)}{\sin(x)} + 1\right)\right]^2\]

Factor out the \(\sin(x)\) in the denominator to get


\[\left[\frac{\sin(x)}{\sin(x)}\left(\cos(x) + \sin(x)\right)\right]^2\]

so now you have just

\[[\cos(x) + \sin(x)]^2\]

Answer 10

But now factor out \(\cos(x)\) to get:


\[\left[\cos(x)\left(1 + \frac{\sin(x)}{\cos(x)}\right)\right]^2\]

Answer 11

Change \(\dfrac{\sin(x)}{\cos(x)}\) to \(\tan(x)\)

\[\left[\cos(x)\left(1 + \tan(x)\right)\right]^2\]

Answer 12

And then re-distribute the squared to get:

\[\cos^2(x)(1 + \tan(x))^2\]

which is the same as \[\cos^2(x)(\tan(x) + 1)^2\]

Answer 13

Thanks so much!