Question-178992

Question Number 178992 by Spillover last updated on 23/Oct/22

Answered by som(math1967) last updated on 23/Oct/22

θ_2 −θ_1 =θ_3 −θ_2 =...=θ_n −θ_(n−1) =d secθ_1 sec θ_2 +sec θ_2 sec θ_1 +...+secθ_(n−1) secθ_n =(1/(cos θ_1 cosθ_2 )) +(1/(cosθ_2 cosθ_3 )) +...+(1/(cosθ_(n−1) cosθ_n )) =(1/(sind))(((sind)/(cos θ_1 cosθ_2 )) +((sind)/(cosθ_2 cosθ_3 )) +...+((sind)/(cosθ_(n−1) cosθ_n ))) =(1/(sind)){((sin(θ_2 −θ_1 ))/(cos θ_1 cosθ_2 )) +((sin(θ_3 −θ_2 ))/(cosθ_2 cosθ_3 )) +...+((sin(θ_n −θ_(n−1) ))/(cosθ_(n−1) cosθ_n ))} =(1/(sind)){((sinθ_2 cosθ_1 )/(cosθ_1 cosθ_2 )) −((sinθ_1 cosθ_2 )/(cosθ_2 cosθ_1 )) +((sinθ_3 cosθ_2 )/(cosθ_2 cosθ_3 ))−((sinθ_2 cosθ_3 )/(cosθ_2 cosθ_3 )) +... +((sinθ_n cosθ_(n−1) )/(cosθ_(n−1) cosθ_n )) −((sinθ_(n−1) cosθ_n )/(cosθ_n cosθ_(n−1) ))} =(1/(sind)){tanθ_2 −tanθ_1 +tanθ_3 −tanθ_2 +...+tanθ_n −tanθ_(n−1) } =((tanθ_n −tanθ_1 )/(sind))

$$\theta_{\mathrm{2}} −\theta_{\mathrm{1}} =\theta_{\mathrm{3}} −\theta_{\mathrm{2}} =…=\theta_{{n}} −\theta_{{n}−\mathrm{1}} ={d} \\ $$$${sec}\theta_{\mathrm{1}} \mathrm{sec}\:\theta_{\mathrm{2}} +\mathrm{sec}\:\theta_{\mathrm{2}} \mathrm{sec}\:\theta_{\mathrm{1}} +…+{sec}\theta_{{n}−\mathrm{1}} {sec}\theta_{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} {cos}\theta_{\mathrm{2}} }\:+\frac{\mathrm{1}}{{cos}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{3}} }\:+…+\frac{\mathrm{1}}{{cos}\theta_{{n}−\mathrm{1}} {cos}\theta_{{n}} } \\ $$$$=\frac{\mathrm{1}}{{sind}}\left(\frac{{sind}}{\mathrm{cos}\:\theta_{\mathrm{1}} {cos}\theta_{\mathrm{2}} }\:+\frac{{sind}}{{cos}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{3}} }\:+…+\frac{{sind}}{{cos}\theta_{{n}−\mathrm{1}} {cos}\theta_{{n}} }\right) \\ $$$$=\frac{\mathrm{1}}{{sind}}\left\{\frac{{sin}\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right)}{\mathrm{cos}\:\theta_{\mathrm{1}} {cos}\theta_{\mathrm{2}} }\:+\frac{{sin}\left(\theta_{\mathrm{3}} −\theta_{\mathrm{2}} \right)}{{cos}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{3}} }\:+…+\frac{{sin}\left(\theta_{{n}} −\theta_{{n}−\mathrm{1}} \right)}{{cos}\theta_{{n}−\mathrm{1}} {cos}\theta_{{n}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{sind}}\left\{\frac{{sin}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{1}} }{{cos}\theta_{\mathrm{1}} {cos}\theta_{\mathrm{2}} }\:−\frac{{sin}\theta_{\mathrm{1}} {cos}\theta_{\mathrm{2}} }{{cos}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{1}} }\right. \\ $$$$\:\:+\frac{{sin}\theta_{\mathrm{3}} {cos}\theta_{\mathrm{2}} }{{cos}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{3}} }−\frac{{sin}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{3}} }{{cos}\theta_{\mathrm{2}} {cos}\theta_{\mathrm{3}} }\:+… \\ $$$$\left.+\frac{{sin}\theta_{{n}} {cos}\theta_{{n}−\mathrm{1}} }{{cos}\theta_{{n}−\mathrm{1}} {cos}\theta_{{n}} }\:−\frac{{sin}\theta_{{n}−\mathrm{1}} {cos}\theta_{{n}} }{{cos}\theta_{{n}} {cos}\theta_{{n}−\mathrm{1}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{sind}}\left\{{tan}\theta_{\mathrm{2}} −{tan}\theta_{\mathrm{1}} +{tan}\theta_{\mathrm{3}} −{tan}\theta_{\mathrm{2}} \right. \\ $$$$\left.+…+{tan}\theta_{{n}} −{tan}\theta_{{n}−\mathrm{1}} \right\} \\ $$$$=\frac{{tan}\theta_{{n}} −{tan}\theta_{\mathrm{1}} }{{sind}} \\ $$

Commented by Spillover last updated on 23/Oct/22

$$\mathrm{very}\:\mathrm{impressive}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by Tawa11 last updated on 23/Oct/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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