How do you prove that $cos 2 x (1 + tan 2 x) = 1$ $cos 2x(1 + tan 2x) = 1$ ?

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How do you prove that $cos 2 x (1 + tan 2 x) = 1$ $cos 2x(1 + tan 2x) = 1$ ?

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Answer 1 · 2015-02-02T11:22:54

The answer2 are: $x = k π$ $x=kpi$ and $\frac{π}{4} + k π$ $pi/4+kpi$ .

The equation can be written:

$cos 2 x + cos 2 x tan 2 x = 1 \Rightarrow cos 2 x + cos 2 x \frac{sin 2 x}{cos 2 x} = 1 \Rightarrow cos 2 x + sin 2 x = 1 \Rightarrow sin 2 x + cos 2 x = 1$ $cos2x+cos2xtan2x=1rArrcos2x+cos2x(sin2x)/(cos2x)=1rArrcos2x+sin2x=1rArrsin2x+cos2x=1$ .

Now it is possible multiply both members for $\frac{\sqrt{2}}{2}$ $sqrt2/2$ :

$\frac{\sqrt{2}}{2} sin 2 x + \frac{\sqrt{2}}{2} cos 2 x = \frac{\sqrt{2}}{2} \Rightarrow$ $sqrt2/2sin2x+sqrt2/2cos2x=sqrt2/2rArr$

$sin 2 x cos (\frac{π}{4}) + cos 2 x sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin2xcos(pi/4)+cos2xsin(pi/4)=sqrt2/2$ .

Using the addition formula:

$sin (2 x + \frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin(2x+pi/4)=sqrt2/2$ .

The sinus is $\frac{\sqrt{2}}{2}$ $sqrt2/2$ if its argument is $\frac{π}{4} + 2 k π$ $pi/4+2kpi$ or $\frac{3}{4} π + 2 k π$ $3/4pi+2kpi$ .

So:

$2 x + \frac{π}{4} = \frac{π}{4} + 2 k π \Rightarrow 2 x = 2 k π \Rightarrow x = k π$ $2x+pi/4=pi/4+2kpirArr2x=2kpirArrx=kpi$ ,

and

$2 x + \frac{π}{4} = \frac{3}{4} π + 2 k π \Rightarrow 2 x = \frac{π}{2} + 2 k π \Rightarrow x = \frac{π}{4} + k π$ $2x+pi/4=3/4pi+2kpirArr2x=pi/2+2kpirArrx=pi/4+kpi$ .

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How do you prove that $cos 2 x (1 + tan 2 x) = 1$ $cos 2x(1 + tan 2x) = 1$ ?

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