Question

If ${\log }_{10}\tan {31}^o.{\log }_{10}\tan {32}^o...{\log }_{10}\tan {60}^o={\log }_{10}a,$then $a=$

Answer 1

Given,

${\log }_{10}a={\log }_{10}\tan {31}^o.{\log }_{10}\tan {32}^o...{\log }_{10}\tan {45}^o...{\log }_{10}\tan {60}^o$

$={\log }_{10}\tan {31}^o...{\log }_{10}1...{\log }_{10}\tan {60}^o$

$=0$

${\log }_{10}a=0$

$a=1$