Question

Sa se arate ca daca triunghiul ABC este dreptunghic in A si cu masura B=75 grade, atunci are loc relatia : tg²B +tg²C=14

Answer 1

tg²B +tg²C= $\frac{sin ^2B}{cos^2B} + \frac{sin ^2C}{cos ^2C} =$

$= \frac{sin ^2(45+30)}{cos^2(45+30)} + \frac{sin ^2(45-30)}{cos ^2(45+30)} =$

$= \frac{(sin 45cos30+sin30cos)^2}{(cos45cos30-sin45sin30)^2} + \frac{(sin 45cos30-sin30cos)^2}{(cos45cos30+sin45sin30)^2} =$

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

$= \frac{( \frac{ \sqrt{6} }{4} + \frac{ \sqrt{2} }{4})^2}{(\frac{ \sqrt{6} }{4}-\frac{ \sqrt{2} }{4})^2} + \frac{(\frac{ \sqrt{6} }{4}- \frac{ \sqrt{2} }{4})^2}{(\frac{ \sqrt{6} }{4}+\frac{ \sqrt{2} }{4})^2} =$

$= \frac{ \frac{1}{16}* ( { \sqrt{6} } + { \sqrt{2} })^2}{\frac{1}{16}* ( { \sqrt{6} } - { \sqrt{2} })^2} + \frac{\frac{1}{16}* ( { \sqrt{6} } - { \sqrt{2} })^2}{\frac{1}{16}* ( { \sqrt{6} } + { \sqrt{2} })^2} =$

$= \frac{( { \sqrt{6} } + { \sqrt{2} })^2}{({ \sqrt{6} }-{ \sqrt{2} })^2} + \frac{({ \sqrt{6} }- { \sqrt{2} })^2}{({ \sqrt{6} }+ \sqrt{2})^2} =$


$= \frac{ 6+2+4 { \sqrt{3}}}{6+2-4{ \sqrt{3} }} + \frac{ 6+2- 4{ \sqrt{3} }}{6+2+ 4\sqrt{3}} =$

$= \frac{ 8+4 { \sqrt{3}}}{8-4{ \sqrt{3} }} + \frac{ 8- 4{ \sqrt{3} }}{8+ 4\sqrt{3}} =$

$= \frac{ 4*(2+ { \sqrt{3}})}{4*(2-{ \sqrt{3} })} + \frac{4*( 2-{ \sqrt{3} })}{4*(2+ \sqrt{3})} =$

$= \frac{ (2+ { \sqrt{3}})}{(2-{ \sqrt{3} })} + \frac{( 2-{ \sqrt{3} })}{(2+ \sqrt{3})} =$

$= \frac{ (2+ { \sqrt{3}})(2+ \sqrt{3})}{(2-{ \sqrt{3} })(2+ \sqrt{3})} + \frac{( 2-{ \sqrt{3} })(2- \sqrt{3})}{(2+ \sqrt{3})(2- \sqrt{3})} =$

$= \frac{ 4+3+ 4{ \sqrt{3}}}{4-3} + \frac{4+3-4{ \sqrt{3} }}{4-3} =$ 

$= 4+3+ 4{ \sqrt{3}} + 4+3-4{ \sqrt{3} } = 7+7= 14$

Answer 2

$\tan^2B+\tan^2C=\tan^2B+cotan^2B=\displaystyle\frac{\sin^2B}{\cos^2B}+\frac{\cos^2B}{\sin^2B}=\\=\displaystyle\frac{\sin^4B+\cos^4B}{\sin^2B\cos^2B}=$
$=\displaystyle\frac{\left(\sin^2B+\cos^2B\right)^2-2\sin^2B\cos^2B}{\sin^2B\cos^2B}=\frac{1-2\sin^2B\cos^2B}{\sin^2B\cos^2B}$  (1)
Avem
$2\sin B\cos B=\sin 2B=\sin 150^{\circ}=\sin(180^{\circ}-150^{\circ})=\sin 30^{\circ}=\frac{1}{2}$
Atunci $\sin B\cos B=\frac{1}{4}$
Înlocuind în (1) și făcând calculele se obține 14.