Proving Identity: Tan(Tan⁻¹(x) + Tan⁻¹(y))

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SUMMARY

The identity proposed, $$\tan[\tan^{-1} (x) +\tan^{-1} (y)] =(x+y)(x-y)$$, is incorrect. The correct formulation is derived using the sum formula for tangent, yielding $$\tan[\tan^{-1} (x) +\tan^{-1} (y)] = \frac{x+y}{1-xy}$$. This was confirmed by testing specific values, such as $$x = y = \tan(\pi/6)$$, which demonstrated the discrepancy in the original identity. The discussion highlights the importance of accurately applying trigonometric identities in proofs.

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karush
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Prove identity

$$\tan[\tan^{-1} (x) +\tan^{-1} (y)] =(x+y)(x-y)$$

Since $$\tan\left({\tan^{-1} \left({a}\right)}\right)=a$$

And by sum formula of $\tan{(x+y)}$ then $$=\frac{x+y}{1-xy}$$

But then?
 
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karush said:
Prove identity

$$\tan[\tan^{-1} (x) +\tan^{-1} (y)] =(x+y)(x-y)$$

Since $$\tan\left({\tan^{-1} \left({a}\right)}\right)=a$$

And by sum formula of $\tan{(x+y)}$ then $$=\frac{x+y}{1-xy}$$

But then?

Hey karush,

It's not true.
Pick for instance $x=y=\tan(\pi/6)$, then $\tan(\pi/6+\pi/6)\ne 0$. :eek:
 
OK, think I see why, should be $$\frac{x+y}{1-xy}$$ in original op
Was reading a very hazy cell phone pic
 
Last edited:

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