Let $$\left(\frac{\cos\alpha}{\sin\alpha}\right) = \left(\frac{b}{a}\right)^n \left(\frac{x_0}{y_0}\right)^\frac{1}{n - 1}$$

Now suppose if I multiply by power $(n - 1)$ to both sides then the equation will look something like this :

$$\left(\frac{\cos\alpha}{\sin\alpha}\right)^\left(n - 1\right) = \left(\frac{b}{a}\right)^n \left(\frac{x_0}{y_0}\right)$$

Have I committed some mistake here? I'm asking this because I'm confused. Also I'm not getting correct answer of my question. I think I have did all the steps correctly. I just have doubt in this one.

The doubt is whether multiplying by power $(n - 1)$ to both sides then l.h.s is easy but how will the $\left(\frac{b}{a}\right)^n$ term on r.h.s will be effected. Kindly point out my mistake.

  • $\begingroup$ you forgot to raise $(b/a)^n$ to the power as well $\endgroup$ – Zach Effman Mar 17 '16 at 16:38
  • $\begingroup$ @ZachEffman where? $\endgroup$ – Hisenberg Mar 17 '16 at 16:41
  • $\begingroup$ In the only step you show here $\endgroup$ – Zach Effman Mar 17 '16 at 16:42
  • $\begingroup$ Also $(\frac{x_0}{y_0})^{(\frac{1}{n-1})^{n-1}} = \frac{x_0}{y_0}$, not what you have $\endgroup$ – Zach Effman Mar 17 '16 at 16:43
  • $\begingroup$ @ZachEffman I don't think so. Kindly see the steps again. $\endgroup$ – Hisenberg Mar 17 '16 at 16:44

In general $(ab)^n$ = $a^n b^n$ and $(a^n)^m = a^{nm}$. Therefore raising both sides to the power $n-1$ yields

$$\left(\frac{\cos\alpha}{\sin\alpha}\right)^{n-1} = \left(\left(\frac{b}{a}\right)^{n} \left(\frac{x_0}{y_0}\right)^\frac{1}{n-1}\right)^{n-1} =\left(\frac{b}{a}\right)^{n(n-1)} \left(\frac{x_0}{y_0}\right).$$

  • $\begingroup$ <a href="math.stackexchange.com/a/1700351/317580">so I think this one's wrong</a> $\endgroup$ – Hisenberg Mar 17 '16 at 16:50
  • $\begingroup$ @user109256 I don't understand what you're saying. $\endgroup$ – Alex Provost Mar 17 '16 at 16:53
  • $\begingroup$ I think this solution has a mistake math.stackexchange.com/a/1700351/317580 see the similar steps. He manipulated them wrong. $\endgroup$ – Hisenberg Mar 17 '16 at 16:55
  • $\begingroup$ @user109256 I don't see a similar manipulation in his answer. $\endgroup$ – Alex Provost Mar 17 '16 at 16:57
  • $\begingroup$ the step where he equated $m_1 = m_2$ $\endgroup$ – Hisenberg Mar 17 '16 at 16:59

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