Another Trig. Identity Question...Any takers?

[math_fortress]

1.) 1/(csc x)(sec^2 x) = ?

a.) 1/(sin x)(cos^2 x)
b.) sin x - sin^3 x
c.) 1/(sin x)(1+tan^2 x)
d.) sin x - (1/1+tan^2 x)
e.) 1+tan^3 x



Here's my work: 1/(csc x)(tan^2 x + 1)
(1/sin x) * ((cos x/sin x)(cos x/sin x) + 1)

So, am I on the right track here? Or do I need to do completely different stuff? I can't seem to find any of those 5 answers no matter what I do...

[nolachrymose]

For (1), rewrite the equation in terms of sine and cosine:

\frac{1}{\csc{x}\sec^2{x}}&=\sin{x}\cos^2{x}

And take it from there (hint: use the basic Pythagorean identity).
However, I'm assuming you meant that original equation is
\displaystyle{\frac{1}{\csc{x}\sec^2{x}}}.

[kreil]

1/(cscx)(sec^2x)
1/(1/sinxcos^2x)

mult. top and bottom by sinxcos^2x

sinxcos^2x

Remember that cos^2x=1-sin^2x
sinx(1-sin^2x)
sinx-sin^3x

so the answer is (b)

[jai6638]

[QUOTE=nolachrymose]For (1), rewrite the equation in terms of sine and cosine:

\frac{1}{\csc{x}\sec^2{x}}&=\sin{x}\cos^2{x}

and then u could say that

Sinx X ( 1 - Sin^2 x) ( because Cos^2 x = 1-Sin^2 x )

therefore, Sinx - Sin^3x ( Answer : B )

[nolachrymose]

Right, that's what I was trying to lead math_fortress to. ;)

[jai6638]

Right, that's what I was trying to lead math_fortress to. ;)

ahh my bad.. im sorry...

[nolachrymose]

No problem. :)