How do I prove $\csc^4 x-\cot^4x=\dfrac{(1+\cos^2x)}{\sin^2x}$

Do you start from RHS or LHS? I get stuck after first few steps-

  • 1
    $\begingroup$ The right-hand side is currently not right. Our function on the left is $\frac{1-\cos^4 x}{\sin^4 x}$. But $1-\cos^4 x=(1-\cos^2 x)(1+\cos^2 x)=\sin^2 x(1+\cos^2 x)$. $\endgroup$ – André Nicolas May 11 '14 at 22:18
  • $\begingroup$ ok I edited the identity, sorry for confusion... $\endgroup$ – Thetasquared May 11 '14 at 22:23
  • $\begingroup$ hint: ${cos^2X}/{sin^2X} = cotX$ $\endgroup$ – Jason Chen May 11 '14 at 22:31
  • $\begingroup$ are you sure it's $1+{cos^2X}/{sin^2X}$ and not ${(1+cos^2X)}/{sin^2X}$? $\endgroup$ – Jason Chen May 11 '14 at 22:33

The LHS: $$\frac{1}{\sin^4 x}-\frac{\cos^4 x}{\sin^4 x}=\frac{1-\cos^4 x}{\sin^4 x}=\frac{(1-\cos^2 x)(1+\cos^2 x)}{\sin^4 x}=\frac{1+\cos^2 x}{\sin^2 x}$$


There is an explanation to your problem at http://www.algebra.com/algebra/homework/Trigonometry-basics.faq.question.43447.html.

  • $\begingroup$ sorry about the ads on the page $\endgroup$ – Jason Chen May 11 '14 at 22:34
  • $\begingroup$ "Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline." - i.e. try from refrain from just adding a link unless the question is a reference request. $\endgroup$ – nbubis May 11 '14 at 23:07

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