PrethodniSledeći
278.

Adicione formule - zbir i razlika dva ugla 4

TEKST ZADATKA

Uprostiti izraz:

tg⁡(x+y)−tg⁡x−tg⁡y\tg{(x+y)}-\tg{x}-\tg{y}
REŠENJE ZADATKA

Primeniti formulu za tangens zbira dva ugla: tg⁡(α+β)=tg⁡α+tg⁡β1−tg⁡αtg⁡β,α=/ π2+πk,β=/ π2+πn,k,n∈Z,tg⁡αtg⁡β=/ 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

tg⁡x+tg⁡y1−tg⁡xtg⁡y−tg⁡x−tg⁡y\frac {\tg{x}+\tg{y}} {1-\tg{x}\tg{y}}-\tg{x}-\tg{y}

Svesti sve činioce izraza na isti imenilac:

tg⁡x+tg⁡y1−tg⁡xtg⁡y−tg⁡x(1−tg⁡xtg⁡y)1−tg⁡xtg⁡y−tg⁡y(1−tg⁡xtg⁡y)1−tg⁡xtg⁡y=tg⁡x+tg⁡y−tg⁡x(1−tg⁡xtg⁡y)−tg⁡y(1−tg⁡xtg⁡y)1−tg⁡xtg⁡y\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:

tg⁡x+tg⁡y−tg⁡x+tg⁡2xtg⁡y−tg⁡y+tg⁡xtg⁡2y1−tg⁡xtg⁡y\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.

tg⁡2xtg⁡y+tg⁡xtg⁡2y1−tg⁡xtg⁡y\frac {\tg^2{x}\tg{y}+\tg{x}\tg^2{y}} {1-\tg{x}\tg{y}}

Izvući zajedničke činioce ispred zagrade:

tg⁡xtg⁡y(tg⁡x+tg⁡y)1−tg⁡xtg⁡y\frac {\tg{x}\tg{y}(\tg{x}+\tg{y})} {1-\tg{x}\tg{y}}

Primeniti formulu za tangens zbira dva ugla: tg⁡(α+β)=tg⁡α+tg⁡β1−tg⁡αtg⁡β,α=/ π2+πk,β=/ π2+πn,k,n∈Z,tg⁡αtg⁡β=/ 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

tg⁡xtg⁡ytg⁡(x+y)\tg{x}\tg{y}\tg{(x+y)}

Da li je rešenje bilo korisno?

Jedan klik nam pomaže da poboljšamo zadatke.

Prijavi se za ocenu