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I need to verify this identity but I have no clue how to solve it. I have tried many different ways but haven't been able to figure it out. $$\frac{\cos^2(t)+\tan^2(t)-1}{\sin^2(t)}=\tan^2(t).$$

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  • $\begingroup$ It is not clear what the expression is. $\endgroup$ – user99914 Apr 5 '15 at 2:15
  • $\begingroup$ The way it's currently written is not an identity. $cos^2(t)-1/sin^2(t)=0 \implies cos^2(t)=1/sin^2(t) \implies cos^2(t)=csc^2(t)$ But this isn't true for all $t$. Can you recheck the expression? $\endgroup$ – suneater Apr 5 '15 at 2:48
  • $\begingroup$ @Brittany: I have rewritten your expression to what I hope you intended. Please correct me if I am wrong. Perhaps you meant to write $(\cos^2 t+\tan^2 t-1)/\sin^2 t$ and left out the parentheses. $\endgroup$ – André Nicolas Apr 5 '15 at 3:32
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$$\dfrac{\cos^2(t)+\tan^2(t)-1}{\sin^2(t)}=\dfrac{\tan^2(t)-\sin^2(t)}{\sin^2(t)}=\dfrac{\tan^2(t)}{\sin^2(t)}-1=\sec^2(t)-1=\tan^2(t).$$

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