Another Trig. Identity Question Any takers?

  • #1
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...
 

Answers and Replies

  • #2
For (1), rewrite the equation in terms of sine and cosine:

[tex]\frac{1}{\csc{x}\sec^2{x}}&=\sin{x}\cos^2{x}[/tex]

And take it from there (hint: use the basic Pythagorean identity).
However, I'm assuming you meant that original equation is
[tex]\displaystyle{\frac{1}{\csc{x}\sec^2{x}}}[/tex].
 
Last edited:
  • #3
kreil
Insights Author
Gold Member
668
68
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)
 
  • #4
263
0
nolachrymose said:
For (1), rewrite the equation in terms of sine and cosine:

[tex]\frac{1}{\csc{x}\sec^2{x}}&=\sin{x}\cos^2{x}[/tex]

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 )
 
  • #5
Right, that's what I was trying to lead math_fortress to. ;)
 
  • #6
263
0
nolachrymose said:
Right, that's what I was trying to lead math_fortress to. ;)

ahh my bad.. im sorry...
 
  • #7
No problem. :)
 

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