Zbir tangensa: tg kvadrat π/12 + 3π/12 + 5π/12

Zapisati izraz na drugačiji način:

\tg^2{\bigg(\frac {3\pi} {12}-\frac {2\pi} {12} \bigg)}+\tg^2{\frac {\pi} {4}}+\tg^2{\bigg(\frac {3\pi} {12}+\frac {2\pi} {12} \bigg)}=\tg^2{\bigg(\frac {\pi} {4}-\frac {\pi} {6} \bigg)}+\tg^2{\frac {\pi} {4}}+\tg^2{\bigg(\frac {\pi} {4}+\frac {\pi} {6} \bigg)}

Uvrstiti vrednosti trigonometrijskih funkcija:

\tg^2{\bigg(\frac {\pi} {4}-\frac {\pi} {6} \bigg)}+1+\tg^2{\bigg(\frac {\pi} {4}+\frac {\pi} {6} \bigg)}

Primeniti formulu za transformaciju zbira i razlike u proizvod: $\tg{\alpha} \pm \tg{\beta}=\frac {\sin{(\alpha \pm \beta)}} {\cos{\alpha}\cos{\beta}} , \alpha\mathrlap{\,/}{=}\frac {\pi} 2(2k+1), \beta\mathrlap{\,/}{=}\frac {\pi} 2(2n+1), k,n \in Z$

Primeniti formulu za zbir i razliku tangensa: $\tg{(\alpha\pm\beta)}=\frac {\tg{\alpha} \pm \tg{\beta}} {1\mp\tg{\alpha}\tg{\beta}} , \alpha\mathrlap{\,/}{=}\frac {\pi} 2+\pi k, \beta\mathrlap{\,/}{=}\frac {\pi} 2+\pi n, k,n \in Z, \tg{\alpha}\tg{\beta}\mathrlap{\,/}{=}1$

\bigg(\frac {\tg{\frac {\pi} {4}} - \tg{\frac {\pi} {6}}} {1+\tg{\frac {\pi} {4}}\tg{\frac {\pi} {6}}}\bigg)^2+1+\bigg(\frac {\tg{\frac {\pi} {4}} + \tg{\frac {\pi} {6}}} {1-\tg{\frac {\pi} {4}}\tg{\frac {\pi} {6}}}\bigg)^2

Uvrstiti vrednosti trigonometrijskih funkcija:

\bigg(\frac {1- \frac {\sqrt{3}} 3} {1+1 \cdot\frac {\sqrt{3}} 3}\bigg)^2+1+\bigg(\frac {1 +\frac {\sqrt{3}} 3} {1-1\cdot\frac {\sqrt{3}} 3}\bigg)^2

Srediti izraz.

\bigg(\frac {\frac {3-\sqrt{3}} 3} {\frac {3+\sqrt{3}} 3}\bigg)^2+1+\bigg(\frac {\frac {3+\sqrt{3}} 3} {\frac {3-\sqrt{3}} 3}\bigg)^2

Osloboditi se dvojnog razlomka:

\bigg(\frac {3(3-\sqrt{3})} {3(3+\sqrt{3})}\bigg)^2+1+\bigg(\frac {3(3+\sqrt{3})} {3(3-\sqrt{3})}\bigg)^2

Skratiti zajedničke činioce:

\bigg(\frac {\cancel{3}(3-\sqrt{3})} {\cancel{3}(3+\sqrt{3})}\bigg)^2+1+\bigg(\frac {\cancel{3}(3+\sqrt{3})} {\cancel{3}(3-\sqrt{3})}\bigg)^2=\bigg(\frac {3-\sqrt{3}} {3+\sqrt{3}}\bigg)^2+1+\bigg(\frac {3+\sqrt{3}} {3-\sqrt{3}}\bigg)^2

Osloboditi se dvojnog razlomka i primeniti formulu za kvadrat binoma : $(a\pm b)^2=a^2\pm2ab+b^2$ :

\frac {(3-\sqrt{3})^2} {(3+\sqrt{3})^2}+1+\frac {(3+\sqrt{3})^2} {(3-\sqrt{3})^2}=\frac {3^2-2\cdot 3 \cdot \sqrt{3}+(\sqrt{3})^2} {3^2+2\cdot 3 \cdot \sqrt{3}+(\sqrt{3})^2}+1+\frac {3^2+2\cdot 3 \cdot \sqrt{3}+(\sqrt{3})^2} {3^2-2\cdot 3 \cdot \sqrt{3}+(\sqrt{3})^2}

Srediti izraz.

\frac {9-6\sqrt{3}+3} {9+6\sqrt{3}+3}+1+\frac {9+6\sqrt{3}+3} {9-6\sqrt{3}+3}=\frac {12-6\sqrt{3}} {12+6\sqrt{3}}+1+\frac {12+6\sqrt{3}} {12-6\sqrt{3}}

Svesti činioce izraza na isti imenilac:

\frac {(12-6\sqrt{3})(12-6\sqrt{3})+(12+6\sqrt{3})(12-6\sqrt{3})+(12+6\sqrt{3})(12+6\sqrt{3})} {(12+6\sqrt{3})(12-6\sqrt{3})}

Osloboditi se zagrada i primeniti formulu za razliku kvadrata: $a^2-b^2=(a-b)(a+b) :$

\frac {12 \cdot 12-12\cdot6\sqrt{3}-6\sqrt{3} \cdot 12+6\sqrt{3}\cdot6\sqrt{3}+12^2-(6\sqrt{3})^2+12 \cdot 12+12\cdot6\sqrt{3}+6\sqrt{3} \cdot 12+6\sqrt{3}\cdot6\sqrt{3}} {12^2-(6\sqrt{3})^2}

Srediti izraz.

\frac {144+108+144-108+144+108} {144-108}=\frac {540} {144-108}=15

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