Question Number 59147 by Pranay last updated on 05/May/19
$$\underset{\mathrm{1}°} {\overset{\mathrm{89}°} {\sum}}\:{log}_{\mathrm{2}} {tan}\:{r}° \\ $$
Answered by ajfour last updated on 05/May/19
$$=\mathrm{log}\:_{\mathrm{2}} \mathrm{1}=\mathrm{0}\:. \\ $$
Answered by tanmay last updated on 05/May/19
$$={ln}_{\mathrm{2}} {tan}\mathrm{1}^{{o}} +{ln}_{\mathrm{2}} {tan}\mathrm{2}^{{o}} +…+{ln}_{\mathrm{2}} {tan}\mathrm{89}^{{o}} \\ $$$$={ln}_{\mathrm{2}} \left({tan}\mathrm{1}^{{o}} {tan}\mathrm{2}^{\mathrm{2}} …{tan}\mathrm{45}^{{o}} …{tan}\mathrm{89}^{{o}} \right) \\ $$$${now}\:{look} \\ $$$${tan}\mathrm{89}^{\mathrm{0}} ={tan}\left(\mathrm{90}^{{o}} −\mathrm{1}^{{o}} \right)={cot}\mathrm{1}^{{o}} \\ $$$${so}\:{tan}\mathrm{1}^{{o}} ×{tan}\mathrm{89}^{{o}} \\ $$$$={tan}\mathrm{1}^{{o}} ×{cot}\mathrm{1}^{{o}} \\ $$$$=\mathrm{1} \\ $$$${similarly}\:{tan}\mathrm{2}^{{o}} ×{tan}\mathrm{87}^{{o}} =\mathrm{1} \\ $$$$… \\ $$$$… \\ $$$${so}\: \\ $$$${ln}_{\mathrm{2}} \left({tan}\mathrm{1}^{{o}} {tan}\mathrm{2}^{{o}} …{tan}\mathrm{89}^{{o}} \right) \\ $$$$={ln}_{\mathrm{2}} \left(\mathrm{1}\right)=\mathrm{0} \\ $$