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1-89-log-2-tan-r-




Question Number 59147 by Pranay last updated on 05/May/19
Σ_(1°) ^(89°)  log_2 tan r°
$$\underset{\mathrm{1}°} {\overset{\mathrm{89}°} {\sum}}\:{log}_{\mathrm{2}} {tan}\:{r}° \\ $$
Answered by ajfour last updated on 05/May/19
=log _2 1=0 .
$$=\mathrm{log}\:_{\mathrm{2}} \mathrm{1}=\mathrm{0}\:. \\ $$
Answered by tanmay last updated on 05/May/19
=ln_2 tan1^o +ln_2 tan2^o +...+ln_2 tan89^o   =ln_2 (tan1^o tan2^2 ...tan45^o ...tan89^o )  now look  tan89^0 =tan(90^o −1^o )=cot1^o   so tan1^o ×tan89^o   =tan1^o ×cot1^o   =1  similarly tan2^o ×tan87^o =1  ...  ...  so   ln_2 (tan1^o tan2^o ...tan89^o )  =ln_2 (1)=0
$$={ln}_{\mathrm{2}} {tan}\mathrm{1}^{{o}} +{ln}_{\mathrm{2}} {tan}\mathrm{2}^{{o}} +…+{ln}_{\mathrm{2}} {tan}\mathrm{89}^{{o}} \\ $$$$={ln}_{\mathrm{2}} \left({tan}\mathrm{1}^{{o}} {tan}\mathrm{2}^{\mathrm{2}} …{tan}\mathrm{45}^{{o}} …{tan}\mathrm{89}^{{o}} \right) \\ $$$${now}\:{look} \\ $$$${tan}\mathrm{89}^{\mathrm{0}} ={tan}\left(\mathrm{90}^{{o}} −\mathrm{1}^{{o}} \right)={cot}\mathrm{1}^{{o}} \\ $$$${so}\:{tan}\mathrm{1}^{{o}} ×{tan}\mathrm{89}^{{o}} \\ $$$$={tan}\mathrm{1}^{{o}} ×{cot}\mathrm{1}^{{o}} \\ $$$$=\mathrm{1} \\ $$$${similarly}\:{tan}\mathrm{2}^{{o}} ×{tan}\mathrm{87}^{{o}} =\mathrm{1} \\ $$$$… \\ $$$$… \\ $$$${so}\: \\ $$$${ln}_{\mathrm{2}} \left({tan}\mathrm{1}^{{o}} {tan}\mathrm{2}^{{o}} …{tan}\mathrm{89}^{{o}} \right) \\ $$$$={ln}_{\mathrm{2}} \left(\mathrm{1}\right)=\mathrm{0} \\ $$

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