278.

Adicione formule

TEKST ZADATKA

Uprostiti izraz:

tg(x+y)tgxtgy\tg{(x+y)}-\tg{x}-\tg{y}

REŠENJE ZADATKA

Primeniti formulu za tangens zbira dva ugla: tg(α+β)=tgα+tgβ1tgαtgβ,α=/π2+πk,β=/π2+πn,k,nZ,tgαtgβ=/1 \tg{(\alpha+\beta)}=\frac {\tg{\alpha}+\tg{\beta}} {1-\tg{\alpha}\tg{\beta}} , \alpha {=}\mathllap{/\,} \frac {\pi} 2 +\pi k, \beta {=}\mathllap{/\,} \frac {\pi} 2+\pi n , k,n \in Z, \tg{\alpha}\tg{\beta}{=}\mathllap{/\,}1

tgx+tgy1tgxtgytgxtgy\frac {\tg{x}+\tg{y}} {1-\tg{x}\tg{y}}-\tg{x}-\tg{y}

Svesti sve činioce izraza na isti imenilac:

tgx+tgy1tgxtgytgx(1tgxtgy)1tgxtgytgy(1tgxtgy)1tgxtgy=tgx+tgytgx(1tgxtgy)tgy(1tgxtgy)1tgxtgy\frac {\tg{x}+\tg{y}} {1-\tg{x}\tg{y}}-\frac {\tg{x}(1-\tg{x}\tg{y})} {1-\tg{x}\tg{y}}-\frac {\tg{y}(1-\tg{x}\tg{y})} {1-\tg{x}\tg{y}}=\frac {\tg{x}+\tg{y}-\tg{x}(1-\tg{x}\tg{y})-\tg{y}(1-\tg{x}\tg{y})} {1-\tg{x}\tg{y}}

Osloboditi se zagrada:

tgx+tgytgx+tg2xtgytgy+tgxtg2y1tgxtgy\frac {\tg{x}+\tg{y}-\tg{x}+\tg^2{x}\tg{y}-\tg{y}+\tg{x}\tg^2{y}} {1-\tg{x}\tg{y}}

Srediti izraz.

tg2xtgy+tgxtg2y1tgxtgy\frac {\tg^2{x}\tg{y}+\tg{x}\tg^2{y}} {1-\tg{x}\tg{y}}

Izvući zajedničke činioce ispred zagrade:

tgxtgy(tgx+tgy)1tgxtgy\frac {\tg{x}\tg{y}(\tg{x}+\tg{y})} {1-\tg{x}\tg{y}}

Primeniti formulu za tangens zbira dva ugla: tg(α+β)=tgα+tgβ1tgαtgβ,α=/π2+πk,β=/π2+πn,k,nZ,tgαtgβ=/1 \tg{(\alpha+\beta)}=\frac {\tg{\alpha}+\tg{\beta}} {1-\tg{\alpha}\tg{\beta}} , \alpha {=}\mathllap{/\,} \frac {\pi} 2 +\pi k, \beta {=}\mathllap{/\,} \frac {\pi} 2+\pi n , k,n \in Z, \tg{\alpha}\tg{\beta}{=}\mathllap{/\,}1

tgxtgytg(x+y)\tg{x}\tg{y}\tg{(x+y)}

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