Proizvod (1 + tg α)(1 + tg β) kada je α + β = π/4

Zapisujemo tangens preko sinusa i kosinusa:

(1 + \text{tg } \alpha)(1 + \text{tg } \beta) = \left(1 + \frac{\sin \alpha}{\cos \alpha}\right)\left(1 + \frac{\sin \beta}{\cos \beta}\right)

Svaku zagradu svodimo na zajednički imenilac:

= \left(\frac{\cos \alpha + \sin \alpha}{\cos \alpha}\right)\left(\frac{\cos \beta + \sin \beta}{\cos \beta}\right)

Množimo razlomke i proširujemo brojilac:

= \frac{\cos \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta}

Koristimo transformacije proizvoda u zbir za sabirke u brojiocu i imeniocu. Prema poznatim formulama imamo:

\begin{aligned} \cos \alpha \cos \beta &= \frac{1}{2}(\cos(\alpha + \beta) + \cos(\alpha - \beta)) \\ \sin \alpha \sin \beta &= \frac{1}{2}(\cos(\alpha - \beta) - \cos(\alpha + \beta)) \\ \sin \alpha \cos \beta &= \frac{1}{2}(\sin(\alpha + \beta) + \sin(\alpha - \beta)) \\ \cos \alpha \sin \beta &= \frac{1}{2}(\sin(\alpha + \beta) - \sin(\alpha - \beta)) \end{aligned}

Zamenjujemo ove izraze u brojilac i grupišemo slične članove:

\text{Brojilac} = \frac{1}{2}(\cos(\alpha + \beta) + \cos(\alpha - \beta)) + \frac{1}{2}(\sin(\alpha + \beta) - \sin(\alpha - \beta)) + \frac{1}{2}(\sin(\alpha + \beta) + \sin(\alpha - \beta)) + \frac{1}{2}(\cos(\alpha - \beta) - \cos(\alpha + \beta))

Sređivanjem brojioca dobijamo:

\text{Brojilac} = \cos(\alpha - \beta) + \sin(\alpha + \beta)

Kako je dato da je $\alpha + \beta = \frac{\pi}{4} ,$ zamenjujemo tu vrednost u brojilac:

\text{Brojilac} = \cos(\alpha - \beta) + \sin\left(\frac{\pi}{4}\right) = \cos(\alpha - \beta) + \frac{\sqrt{2}}{2}

Sada primenjujemo formulu za proizvod kosinusa na imenilac i zamenjujemo $\alpha + \beta = \frac{\pi}{4} :$

\text{Imenilac} = \cos \alpha \cos \beta = \frac{1}{2}(\cos(\alpha + \beta) + \cos(\alpha - \beta)) = \frac{1}{2}\left(\cos\left(\frac{\pi}{4}\right) + \cos(\alpha - \beta)\right) = \frac{1}{2}\left(\frac{\sqrt{2}}{2} + \cos(\alpha - \beta)\right)

Vraćamo sređeni brojilac i imenilac u polazni razlomak:

\frac{\cos(\alpha - \beta) + \frac{\sqrt{2}}{2}}{\frac{1}{2}\left(\frac{\sqrt{2}}{2} + \cos(\alpha - \beta)\right)}

Skraćujemo razlomak tako što podelimo iste izraze u brojiocu i imeniocu:

= \frac{\cos(\alpha - \beta) + \frac{\sqrt{2}}{2}}{\frac{1}{2}\left(\cos(\alpha - \beta) + \frac{\sqrt{2}}{2}\right)} = \frac{1}{\frac{1}{2}} = 2

Ovim smo pokazali da je početni izraz zaista jednak 2.

(1 + \text{tg } \alpha)(1 + \text{tg } \beta) = 2

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