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Question Number 120621 by TANMAY PANACEA last updated on 01/Nov/20

$${Trigonometry} \\ $$$${selective}\:{questions} \\ $$

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

S=(1+tan1°)(1+tan2°)... (1+tan1°)=1+((1−tan44°)/(1+tan44°))=(2/(1+tan44°)) (1+tan2°)=(2/(1+tan43°)) S^2 =4.(2.2.2...44) S^2 =2^(46) S=2^(23) n=23

$${S}=\left(\mathrm{1}+{tan}\mathrm{1}°\right)\left(\mathrm{1}+{tan}\mathrm{2}°\right)… \\ $$$$\left(\mathrm{1}+{tan}\mathrm{1}°\right)=\mathrm{1}+\frac{\mathrm{1}−{tan}\mathrm{44}°}{\mathrm{1}+{tan}\mathrm{44}°}=\frac{\mathrm{2}}{\mathrm{1}+{tan}\mathrm{44}°} \\ $$$$\left(\mathrm{1}+{tan}\mathrm{2}°\right)=\frac{\mathrm{2}}{\mathrm{1}+{tan}\mathrm{43}°} \\ $$$${S}^{\mathrm{2}} =\mathrm{4}.\left(\mathrm{2}.\mathrm{2}.\mathrm{2}…\mathrm{44}\right) \\ $$$${S}^{\mathrm{2}} =\mathrm{2}^{\mathrm{46}} \\ $$$${S}=\mathrm{2}^{\mathrm{23}} \\ $$$${n}=\mathrm{23} \\ $$

Commented by TANMAY PANACEA last updated on 01/Nov/20

(1+tan1) (tan45+tan1)=(((sin45cos1+cos45sin1)/(cos45cos1))) (((sin46)/(cos45cos1))) so p=((sin46)/(cos45cos1))×((sin47)/(cos45cos2))×((sin48)/(cos45cos3))×..×((sin89)/(cos45cos44))×((sin90)/(cos45cos45)) =now look sin46=cos44.. sin47=cos43 so cancel out p=((1/(cos45)))^(44) ×((1/(cos45)))^2 =((√2) )^(46) =2^(23)

$$\left(\mathrm{1}+{tan}\mathrm{1}\right) \\ $$$$\left({tan}\mathrm{45}+{tan}\mathrm{1}\right)=\left(\frac{{sin}\mathrm{45}{cos}\mathrm{1}+{cos}\mathrm{45}{sin}\mathrm{1}}{{cos}\mathrm{45}{cos}\mathrm{1}}\right) \\ $$$$\left(\frac{{sin}\mathrm{46}}{{cos}\mathrm{45}{cos}\mathrm{1}}\right) \\ $$$${so} \\ $$$${p}=\frac{{sin}\mathrm{46}}{{cos}\mathrm{45}{cos}\mathrm{1}}×\frac{{sin}\mathrm{47}}{{cos}\mathrm{45}{cos}\mathrm{2}}×\frac{{sin}\mathrm{48}}{{cos}\mathrm{45}{cos}\mathrm{3}}×..×\frac{{sin}\mathrm{89}}{{cos}\mathrm{45}{cos}\mathrm{44}}×\frac{{sin}\mathrm{90}}{{cos}\mathrm{45}{cos}\mathrm{45}} \\ $$$$={now}\:{look} \\ $$$${sin}\mathrm{46}={cos}\mathrm{44}.. \\ $$$${sin}\mathrm{47}={cos}\mathrm{43}\:\:{so}\:{cancel}\:{out} \\ $$$${p}=\left(\frac{\mathrm{1}}{{cos}\mathrm{45}}\right)^{\mathrm{44}} ×\left(\frac{\mathrm{1}}{{cos}\mathrm{45}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{46}} =\mathrm{2}^{\mathrm{23}} \\ $$

Commented by TANMAY PANACEA last updated on 01/Nov/20