Question

tg 25° . tg 65° = 1.

Answer 1

 

  $\displaystyle\bf\\tg~25^o\times tg~65^o=?\\\\Folosim~formulele:\\\\tg~x=\frac{sin~x}{cos~x}\\\\sin(90^o-x)=cos~x \\\\cos(90^o-x)=sin~x\\\\Rezolvare\!:\\\\tg~25^o\times tg~65^o=\\\\=\frac{sin~25}{cos~25}\times\frac{sin~65}{cos~55}=\\\\\\=\frac{sin~25}{cos~25}\times\frac{sin(90^o-25^o)}{cos(90^o-25^o)}=\\\\\\=\frac{sin~25}{cos~25}\times\frac{cos~25^o}{sin~25^o}=\\\\\\=\frac{sin~25\times cos~25^o}{cos~25\times sin~25}=\boxed{\bf1}~~~(Dupa~simplificare)$