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I need to show that $\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}} = \sec\theta - \tan\theta$

So far, I have changed the fraction to (where $t=\tan\frac{\theta}{2}$), $\frac{1-2t+t^2}{1-t^2}$(*)

I know that $\tan\theta = \frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}$ and that I should be able to use this to complete the proof, but I am unsure how to split the fraction (*) I have arrived at in order to utilise this?

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$$\frac{1-\tan \frac x2}{1+\tan \frac x2}=\frac{\cos \frac x2-\sin \frac x2}{\cos \frac x2+\sin \frac x2}=\frac{1-\sin x}{\cos x}=\sec x-\tan x.$$ $\hspace{4cm}$ (Change to $\sin$ and $\cos$)-(Multiply $N^m$ and $D^m$ by $N^m$)

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Notice, $$\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}=\frac{\left(1-\tan\frac{\theta}{2}\right)\left(1+\tan\frac{\theta}{2}\right)}{\left(1+\tan\frac{\theta}{2}\right)^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}+2\tan\frac{\theta}{2}}$$ $$=\frac{1}{\frac{1+\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}+\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}}=\frac{1}{\sec \theta+\tan \theta}=\frac{\sec\theta-\tan \theta}{\sec^2 \theta-\tan^2\theta}=\sec\theta-\tan \theta$$

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  • $\begingroup$ I don't think the final equality is correct? $\endgroup$ – MathsUndergrad Oct 18 '15 at 10:59
  • $\begingroup$ @MathsUndergrad Remember that $\sec^2\theta - \tan^2\theta = 1$. $\endgroup$ – N. F. Taussig Oct 18 '15 at 11:05
  • $\begingroup$ Ahh yes of course $\endgroup$ – MathsUndergrad Oct 18 '15 at 11:10
  • $\begingroup$ @MathsUndergrad: Is there something incorrect.? $\endgroup$ – Harish Chandra Rajpoot Oct 18 '15 at 11:12
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$$\csc2y-\cot2y=\dfrac{1-\cos2y}{\sin2y}=\cdots=\tan y$$

Now set $y=\dfrac\pi4-x\iff2y=\dfrac\pi2-2x$

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