Question

tan 15 .tan25+tan15.tan50+tan25.tan50​

Answer 1

CONCEPT:

$\bf\,tan(A+B)=\displaystyle\frac{tanA+tanB}{1-tanA\,tanB}$

consider,

$tan(15+25+50)$

$=\displaystyle\frac{tan(15+25)+tan50}{1-tan(15+25)\,tan50}$

$=\displaystlye\frac{\frac{tan15+tan25}{1-tan15\,tan25}+tan50}{1-\frac{tan15+tan25}{1-tan15\,tan25}\,tan50}$

$=\displaystlye\frac{tan15+tan25+tan50-tan15\,tan25\,tan50}{1-(tan15\,tan25+tan25\,tan50+tan50\,tan15)}$

$\implies\displaystyle\,tan90=\frac{tan15+tan25+tan50-tan15\,tan25\,tan50}{1-(tan15\,tan25+tan25\,tan50+tan50\,tan15)}$

$\implies\displaystlye\frac{tan15+tan25+tan50-tan15\,tan25\,tan50}{1-(tan15\,tan25+tan25\,tan50+tan50\,tan15)}=\infty$

$\implies\,1-(tan15\,tan25+tan25\,tan50+tan50\,tan15)=0$

$\implies\boxed{\bf\,tan15\,tan25+tan25\,tan50+tan50\,tan15=1}$