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Here is a question I am stuck on

Show that $$\tan^2\left(\frac{\pi}{4}+x\right)=\frac{1+\sin(2x)}{1-\sin(2x)}$$

So far, I have been able to simplify to $$\frac{\cos^2x+\sin^2x}{\cos^2x-\sin^2x}$$ But I don't know where to go from here.

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  • $\begingroup$ Which side did you simplify? $\endgroup$ – J.G. Jun 22 '19 at 13:06
  • $\begingroup$ @J.G. Simplifyed LHS first $\endgroup$ – Alex Jun 22 '19 at 13:07
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$\displaystyle \tan^2(\frac{\pi}4+x) \\= (\frac{\tan \frac{\pi}4 + \tan x}{1 - \tan \frac{\pi}4\cdot \tan x})^2 \\ = \frac{(1+\tan x)^2}{(1-\tan x)^2} \\ = \frac {(1+\frac{\sin x}{\cos x})^2}{(1-\frac{\sin x}{\cos x})^2} \\ = \frac{(\cos x + \sin x)^2}{(\cos x - \sin x)^2}\\ = \frac{\sin^2 x + \cos^2x + 2\sin x \cos x}{\sin^2 x + \cos^2x - 2\sin x \cos x} \\ = \frac{1+ 2\sin x\cos x}{1 - 2\sin x\cos x}\\= \frac{1 + \sin 2x}{1 - \sin 2x}$

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$$1\pm\sin2x=\left(\cos x\pm\sin x\right)^2$$

$$\dfrac{\cos x+\sin x}{\cos x+\sin x}=\dfrac{1+\tan x}{1-\tan x}=\tan\left(x+\dfrac\pi4\right)$$

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Hint:

$$\frac{1+\sin 2x}{1-\sin 2x} = \frac{(\sin x + \cos x)^2}{(\sin x - \cos x)^2}$$

$$\tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B}$$

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From RHS

$$\sin(2x) = -\cos\bigg(\frac{\pi}{2}+2x\bigg) = -\cos\bigg(2\big(\frac{\pi}{4}+x\big)\bigg)$$

and use

$$\frac{1-\cos(2\alpha)}{1+\cos(2\alpha)}=\tan^2\alpha$$ to get LHS


Or from LHS directly use the above identity to get $$\tan^2\alpha = \frac{1-\cos(2\alpha)}{1+\cos(2\alpha)}$$

Here $\alpha = x+\frac{\pi}{4}$ and $\cos(2\frac{\pi}{4}+2x) = \cos(\frac{\pi}{2}+2x) = -\sin(2x)$

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$$\begin{align*} \tan^2\left(\frac\pi4+x\right) &= \sec^2\left(\frac\pi4+x\right)-1\\ &= \frac1{\cos^2\left(\frac\pi4+x\right)}-1\\ &= \frac2{2\cos^2\left(\frac\pi4+x\right)}-1\\ &= \frac2{1+2\cos^2\left(\frac\pi4+x\right)-1}-1\\ &= \frac2{1+\cos2\left(\frac\pi4+x\right)}-1\\ &= \frac2{1-\sin2x}-1\\ &= \frac{1+\sin2x}{1-\sin2x}\\ \end{align*}$$

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