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Question-63769

Question Number 63769 by aliesam last updated on 08/Jul/19 Commented by mathmax by abdo last updated on 09/Jul/19 $${let}\:{prove}\:{by}\:{recurence}\:{n}=\mathrm{0}\:\:\:\:{A}_{\mathrm{0}} =\mathrm{0}\:{is}\:{divisible}\:{by}\:\mathrm{6} \\ $$$${let}\:{suppose}\:{A}_{{n}} ={n}^{\mathrm{3}} \:+\mathrm{5}{n}\:{is}\:{divisible}\:{by}\:\mathrm{6}\:\Rightarrow \\…

Question-63763

Question Number 63763 by aliesam last updated on 08/Jul/19 Commented by Prithwish sen last updated on 08/Jul/19 $$\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:+\:\mathrm{i}\:\mathrm{sin}\left(\mathrm{2x}\right)}\:=\:\frac{\mathrm{2}\left\{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)\right\}}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\:+\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)}{\mathrm{1}+\:\mathrm{cos}\left(\mathrm{2x}\right)}\:=\:\mathrm{1}\:−\:\mathrm{i}\frac{\mathrm{2sinx}.\mathrm{cosx}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{1}\:−\mathrm{itanx}\:\mathrm{Hence}\:\mathrm{proved}.…

Tanmay-Sir-Are-you-ok-

Question Number 63758 by Prithwish sen last updated on 08/Jul/19 $$\mathrm{Tanmay}\:\mathrm{Sir}.\:\mathrm{Are}\:\mathrm{you}\:\mathrm{ok}\:? \\ $$ Commented by mathmax by abdo last updated on 09/Jul/19 $${i}\:{see}\:{that}\:{sir}\:{Tanmay}\:{is}\:{absent}\:{in}\:{the}\:{forum}\:.{i}\:{hope}\:{that}\:{we} \\ $$$${know}\:{the}\:{cause}\:{of}\:{his}\:{absence}….…

Q-129271-please-answer-

Question Number 129292 by AgnibhoMukhopadhyay last updated on 14/Jan/21 $$\mathrm{Q}\:\mathrm{129271}\:\mathrm{please}\:\mathrm{answer} \\ $$ Commented by prakash jain last updated on 14/Jan/21 $$\mathrm{I}\:\mathrm{get}\:−\mathrm{ve}\:\mathrm{answer}.\:\mathrm{check}\:\mathrm{original} \\ $$$$\mathrm{questions}\:\mathrm{comment}. \\ $$$$\mathrm{Maybe}\:\mathrm{somebody}\:\mathrm{else}\:\mathrm{can}\:\mathrm{comment}…