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I did it by converting every trigonometry stuff into ${\sin}$ and $\cos$. But I want to know if there is a shortcut (without converting everything to $\sin$ and $\cos$) to do this. Please help.

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marked as duplicate by Arnaud D., lab bhattacharjee trigonometry Jul 15 '16 at 16:37

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  • $\begingroup$ Use two $ signs to enclose your formulas... $\endgroup$ – DonAntonio Jul 15 '16 at 15:35
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    $\begingroup$ Try break all terms in form of $\cos\theta$ and $\sin\theta$... $\endgroup$ – tatan Jul 15 '16 at 15:38
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HINT:

$$\dfrac1{\cot\theta}=\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\sec\theta}{\csc\theta}$$

$$\dfrac{\tan\theta}{1-\cot\theta}=\dfrac{\sec^2\theta}{\csc\theta(\sec\theta-\csc\theta)}$$

Similarly, $$\dfrac{\cot\theta}{1-\tan\theta}=?$$


Alternatively,

$$\dfrac{\tan\theta}{1-\cot\theta}=\dfrac{\sin^2\theta}{\cos\theta(\cos\theta-\sin\theta)}$$

Similarly, $$\dfrac{\cot\theta}{1-\tan\theta}=?$$

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  • $\begingroup$ I got the other part as $\frac{\csc^2\theta}{\sec\theta(\csc\theta-\sec\theta)}$ so what then ? $\endgroup$ – Nimantha Jul 15 '16 at 15:54
  • $\begingroup$ @Nimantha, Now the two terms & simplify. You can use either of the methods $\endgroup$ – lab bhattacharjee Jul 15 '16 at 15:56
  • $\begingroup$ I don't see way to prove what is in the question.Please help. $\endgroup$ – Nimantha Jul 15 '16 at 15:57
  • $\begingroup$ @Nimantha, $$\dfrac{a^2}{b(a-b)}+\dfrac{b^2}{a(a-b)}=\dfrac{a^3-b^3}{ab(a-b)}=\dfrac{a^2+b^2+ab}{ab}=\dfrac{a^2b^2+ab}{ab}=?$$ as $a^2+b^2=\cdots=a^2b^2$ $\endgroup$ – lab bhattacharjee Jul 15 '16 at 16:03
  • $\begingroup$ @Nimantha, please try with sine, cosine : that's much easier to deal with $\endgroup$ – lab bhattacharjee Jul 15 '16 at 16:07

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