Question

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Answer 1

$1)\frac{tg^{2}25^{0}-tg^{2}15^{0}}{1-tg^{2}25^{0}*tg^{2}15^{0}} = \frac{tg25^{0}-tg15^{0}}{1+tg25^{0}*tg15^{0}}*\frac{tg25^{0}+tg15^{0}}{1-tg25^{0}*tg15^{0}}=\\\\=tg(25^{0} -15^{0})*tg(25^{0}+15^{0})=\boxed{tg10^{0}*tg40^{0}}\\\\\\2)\frac{tg^{2}1,8^{0}-tg^{2}1,2^{0}}{1-tg^{2}1,8^{0}*tg^{2}1,2^{0}} =\frac{tg1,8^{0}-tg1,2^{0}}{1+tg1,8^{0}*tg1,2^{0}}*\frac{tg1,8^{0}+tg1,2^{0}}{1-tg1,8^{0}*tg1,2^{0}}=\\\\=tg(1,8^{0}-1,2^{0})*tg(1,8^{0} +1,2^{0})=\boxed{tg0,6^{0}*tg3^{0}}$

$3)tg(\alpha+\beta)*tg(\alpha-\beta)*(1-tg^{2}\alpha*tg^{2}\beta)=\\\\=\frac{tg\alpha+tg\beta}{1-tg\alpha*tg\beta} *\frac{tg\alpha-tg\beta}{1+tg\alpha*tg\beta}*(1-tg^{2}\alpha*tg^{2} \beta)=\frac{tg^{2}\alpha-tg^{2}\beta}{1-tg^{2}\alpha*tg^{2}\beta}*(1-tg^{2}\alpha*tg^{2} \beta)=\\\\=\boxed{tg^{2}\alpha-tg^{2}\beta}$

Answer 2

$1)\ \ \dfrac{tg^225^\circ -tg^215^\circ }{1-tg^225^\circ \cdot tg^215^\circ }=\dfrac{\dfrac{sin^225}{cos^225}-\dfrac{sin^215}{cos^215}}{1-\dfrac{sin^225\cdot sin^215}{cos^225\cdot cos^215}}=\dfrac{sin^225\cdot cos^215-sin^215\cdot cos^225}{cos^225\cdot cos^215-sin^225\cdot sin^215}=$

$=\dfrac{(sin25\cdot cos15-sin15\cdot cos25)(sin25\cdot cos15+sin15\cdot cos25)}{(cos25\cdot cos15-sin25\cdot sin15)(cos25\cdot cos15+sin25\cdot sin15)}=\\\\\\=\dfrac{sin(25-15)\cdot sin(25+15)}{cos(25+15)\cdot cos25-15)}=\dfrac{sin10\cdot sin40}{cos40\cdot cos10}=tg10\cdot tg40$

$2)\ \ \dfrac{tg^21,8-tg^21,2}{1-tg^21,8\cdot tg^21,2}=\dfrac{(tg1,8-tg1,2)(tg1,8+tg1,2)}{(1-tg1,8\cdot tg1,2)(1+tg1,8\cdot tg1,2)tg}=\\\\\\=\dfrac{tg1,8-tg1,2}{1+tg1,8\cdot tg1,2}\cdot \dfrac{tg1,8+tg1,2}{1-tg1,8\cdot tg1,2}=tg(1,8-1,2)\cdot tg(1,8+1,2)=\\\\\\=tg0,6\cdot tg3\qquad \qquad ili$$2)\ \ \dfrac{tg^21,8-tg^21,2}{1-tg^21,8\cdot tg^21,2}=\dfrac{\dfrac{sin^21,8}{cos^21,8}-\dfrac{sin^21,2}{cos^21,2}}{1-\dfrac{sin^21,8\cdot sin^21,2}{cos^21.8\cdot cos^21,2}}=\dfrac{sin^21,8\cdot cos^21,2-sin^21,2\cdot cos^21,8}{cos^21,8\cdot cos^21,2-sin^21,8\cdot sin^21,2}=$

$=\dfrac{(sin1,8\cdot cos1,2-sin1,2\cdot cos1,8)(sin21,8\cdot cos15,2+sin1,2\cdot cos1,8)}{(cos1,8\cdot cos1,2-sin1,8\cdot sin1,2)(cos1,8\cdot cos1,2+sin1,8\cdot sin1,2)}=\\\\\\=\dfrac{sin(1,8-1,2)\cdot sin(1,8+1,2)}{cos(1,8+1,2)\cdot cos1,8-1,2)}=\dfrac{sin0,6\cdot sin3}{cos3\cdot cos0,6}=tg0,6\cdot tg3$

$3)\ \ tg(\alpha +\beta )\cdot tg(\alpha -\beta )\cdot (1-tg^2\alpha \cdot tg^2\beta )=\\\\\\=\dfrac{tg\alpha +tg\beta }{1-tg\alpha \cdot tg\beta }\cdot \dfrac{tg\alpha -tg\beta }{1+tg\alpha \cdot tg\beta }\cdot (1-tg\alpha \cdot tg\beta )(1+tg\alpha \cdot tg\beta )=\\\\\\=(tg\alpha +tg\beta )\cdot (tg\alpha -tg\beta )=tg^2\alpha -tg^2\beta$