# Prove $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=\sec \theta\csc\theta+1$ [duplicate]

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.

## marked as duplicate by Arnaud D., lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 15 '16 at 16:37

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• Use two $signs to enclose your formulas... – DonAntonio Jul 15 '16 at 15:35 • Try break all terms in form of$\cos\theta$and$\sin\theta$... – tatan Jul 15 '16 at 15:38 ## 1 Answer 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, $$\dfrac{\cot\theta}{1-\tan\theta}=?$$ • I got the other part as$\frac{\csc^2\theta}{\sec\theta(\csc\theta-\sec\theta)}$so what then ? – Nimantha Jul 15 '16 at 15:54 • @Nimantha, Now the two terms & simplify. You can use either of the methods – lab bhattacharjee Jul 15 '16 at 15:56 • I don't see way to prove what is in the question.Please help. – Nimantha Jul 15 '16 at 15:57 • @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\$ – lab bhattacharjee Jul 15 '16 at 16:03
• @Nimantha, please try with sine, cosine : that's much easier to deal with – lab bhattacharjee Jul 15 '16 at 16:07