Question

Solve $lo{g}_{10}tan{1}^{\circ }+lo{g}_{10}tan{2}^{\circ }+........+lo{g}_{10}tan{89}^{\circ }$

Answer 1

$(lo{g}_{10}tan{1}^{\circ }+lo{g}_{10}tan{2}^{\circ }+........+lo{g}_{10}tan{89}^{\circ })$
$=lo{g}_{10}(tan{1}^{\circ }tan{2}^{\circ }.....tan{44}^{\circ }tan{45}^{\circ }tan{46}^{\circ }...........tan{89}^{\circ })$
$=lo{g}_{10}\left[\right(tan{1}^{\circ }tan{89}^{\circ }\left)\right(tan{2}^{\circ }tan{88}^{\circ })............(tan{44}^{\circ }tan{46}^{\circ }\left)tan{45}^{\circ }\right]$
$=lo{g}_{10}\left[\right(tan{1}^{\circ }cot{1}^{\circ }\left)\right(tan{2}^{\circ }cot{2}^{\circ })..........(tan{44}^{\circ }cot{44}^{\circ }\left)tan{45}^{\circ }\right]$
$=lo{g}_{10}1=0$
$[\because tan{1}^{\circ }cot{1}^{\circ }=tan{2}^{\circ }cot{2}^{\circ }=...=tan{44}^{\circ }cot{44}^{\circ }=1andtan{45}^{\circ }=1]$