Further simplify tan⁡(α+β)−tan⁡(β)but I keep wondering, whether it's possible to further simplify this, into for example only using the tan⁡ once. I tried using the addition theorems for trigonometry, but these just seem to complicate them further.I already tried something along this:r=tan⁡(α+β)−tan⁡(β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β−tan⁡β=tan⁡α+tan⁡β1−tan⁡αtan⁡β−(1−tan⁡αtan⁡β)(tan⁡β)1−tan⁡αtan⁡β=(tan⁡α+tan⁡β)−(tan⁡β−tan⁡αtan2β)1−tan⁡αtan⁡β=tan⁡α+tan⁡αtan2β1−tan⁡αtan⁡β

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Further simplify tan⁡(α+β)−tan⁡(β)but I keep wondering, whether it&#039;s possible to further simplify this, into for example only using the tan⁡ once. I tried using the addition theorems for trigonometry, but these just seem to complicate them further.I already tried something along this:r=tan⁡(α+β)−tan⁡(β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β−tan⁡β=tan⁡α+tan⁡β1−tan⁡αtan⁡β−(1−tan⁡αtan⁡β)(tan⁡β)1−tan⁡αtan⁡β=(tan⁡α+tan⁡β)−(tan⁡β−tan⁡αtan2β)1−tan⁡αtan⁡β=tan⁡α+tan⁡αtan2β1−tan⁡αtan⁡β

cineworld93uowb · Accepted Answer

Continuing from where you left:tan⁡α+tan⁡αtan2β1−tan⁡αtan⁡β=tan⁡α(1+tan2β)1−tan⁡αtan⁡βNow 1+tan2β=sec2β=1cos2β so we have:tan⁡α(1+tan2β)1−tan⁡αtan⁡β=sin⁡αcos⁡αcos2β1−sin⁡αsin⁡βcos⁡αcos⁡β=sin⁡αcos⁡β(cos⁡αcos⁡β−sin⁡αsin⁡β)Now we know, cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)So you have:sin⁡(α)cos⁡(β)cos⁡(α+β)

jmroberts70pbo2 · Accepted Answer

The expressiontan⁡(α+β)−tan⁡(β)goes to infinity for α+β=dπ2+mπ and β=dπ2+nπAssuming it can be expressed as a fraction, the denominator must have roots at these values, i.e. have the factors cos⁡(α+β)cos⁡(β)The numerator is obviously not a constant, so that it doesn&#039;t seem possible to simplify, i.e. to express the same quantity with less than two trigonometric functions.

Further simplify \(\displaystyle{\tan{{\left(\alpha+\beta\right)}}}-{\tan{{\left(\beta\right)}}}\) but I keep wondering, whether

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