First of all, we construct an Isosceles Triangle like this:

With \(AB=BC=a\) (say). And \(B=2x\). Then \(AD \perp BC\) is constructed.

Subsequently, since the angles \(BAC\) and \(ACB\) are equal, angle \(ACB = 90 - x\), then angle \(CAD=x\).

From triangle \(ADC\):

\(\tan \ x\ =\ \frac{CD}{AD}\)

\(\implies \tan \ x\ =\ \frac{a-a\cos 2x}{a\sin 2x}\)

\(\implies \tan \ x\ =\ \frac{1-\cos 2x}{\sin 2x}\)

\(\implies \cos 2x=1-\sin 2x\tan x\)

Squaring both sides,

\(\implies \cos ^22x=1-2\sin 2x\tan x+\sin ^22x\tan ^2x\)

\(\implies 1-\sin ^22x=1-2\sin 2x\tan x+\sin ^22x\tan ^2x\)

\(\implies \sin ^22x=2\sin 2x\tan x-\sin ^22x\tan ^2x\)

\(\sin 2x\) is not **necessarily** \(0\), so we can divide both sides by \(\sin 2x\).

\(\implies \sin 2x=2\tan x-\sin 2x\tan ^2x\)

\(\implies \sin 2x\left(1+\tan ^2x\right)=2\tan x\)

\(\implies \boxed{\sin 2x=\frac{2\tan x}{1+\tan ^2x}}\)

Furthermore,

\(\sin 2x=\frac{2\tan x}{1+\tan ^2x}=\frac{2\tan x}{\sec ^2x}=2\tan x\cos ^2x=2\frac{\sin x}{\cos x}\cos ^2x=\boxed{2\sin x\cos x}\)

Now, from \(\tan \ x\ =\ \frac{1-\cos 2x}{\sin 2x}\):

\(\sin 2x=\ \frac{1-\cos 2x}{\tan x}\)

\(\implies \frac{2\tan x}{1+\tan ^2x}=\ \frac{\left(1-\cos 2x\right)}{\tan x}\)

\(\implies \frac{2\tan ^2x}{1+\tan ^2x}=\ 1-\cos 2x\)

\(\implies \cos 2x=1-\frac{2\tan ^2x}{1+\tan ^2x}\)

\(\implies \boxed{\cos 2x=\frac{1-\tan ^2x}{1+\tan ^2x}}\)

From here,

\(\cos 2x=\frac{1-\tan ^2x}{1+\tan ^2x} = \frac{\cos ^2x\left(1-\tan ^2x\right)}{\cos ^2x\left(1+\tan ^2x\right)} = \frac{\cos ^2x-\sin ^2x}{\cos ^2x+\sin ^2x} = \frac{\cos ^2x-\sin ^2x}{1} = \boxed{\cos ^2x-\sin ^2x}\)

Combining the expansions,

\(\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\left(\frac{2\tan x}{1+\tan ^2x}\right)}{\left(\frac{1-\tan ^2x}{1+\tan ^2x}\right)} = \boxed{\frac{2\tan x}{1-\tan ^2x}}\)

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## Comments

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TopNewestAll is to confusing

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Yeah, it is but an indirect way of deriving the identities using just basic trigonometry.

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i can not understand anything. explanation anyone??????

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What did you not understand, the construction of and in the triangle?

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