Menu Close

Author: Tinku Tara

3-2-log-9-25-5-9-2-x-

Question Number 4515 by love math last updated on 04/Feb/16 $$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{5}×\mathrm{9}^{\frac{\mathrm{2}}{{x}}} \\ $$ Commented by love math last updated on 05/Feb/16 $${Is}\:{there}\:{any}\:{way}\:{to}\:{prove}\:\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}}…

a-b-c-cos-a-b-2-cos-c-2-

Question Number 70051 by Aditya789 last updated on 30/Sep/19 $$\frac{{a}+{b}}{{c}}=\frac{{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}{{cos}\frac{{c}}{\mathrm{2}}} \\ $$ Answered by $@ty@m123 last updated on 01/Oct/19 $${LHS}=\frac{{a}+{b}}{{c}}=\frac{\mathrm{sin}\:{A}+\mathrm{sin}\:{B}}{\mathrm{sin}\:{C}} \\ $$$$=\frac{\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}}{\mathrm{2sin}\:\frac{{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{C}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{sin}\:\frac{\pi−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{C}}{\mathrm{2}}} \\…

2016-2007-2018-2009-2020-2011-2022-

Question Number 70044 by naka3546 last updated on 30/Sep/19 $$\sqrt{\mathrm{2016}\:+\:\mathrm{2007}\sqrt{\mathrm{2018}\:+\:\mathrm{2009}\sqrt{\mathrm{2020}\:+\:\mathrm{2011}\sqrt{\mathrm{2022}\:+\:\ldots}}}}\:\:=\:\:… \\ $$ Commented by Prithwish sen last updated on 30/Sep/19 $$\left(\mathrm{a}−\mathrm{6}\right)=\sqrt{\left(\mathrm{a}−\mathrm{6}\right)^{\mathrm{2}} }=\sqrt{\mathrm{a}+\left(\mathrm{a}^{\mathrm{2}} −\mathrm{13a}+\mathrm{36}\right)} \\ $$$$\:\:=\:\sqrt{\mathrm{a}+\left(\mathrm{a}−\mathrm{9}\right)\left(\mathrm{a}−\mathrm{4}\right)}\:=\:\sqrt{\mathrm{a}+\left(\mathrm{a}−\mathrm{9}\right)\sqrt{\left(\mathrm{a}−\mathrm{4}\right)^{\mathrm{2}}…

lets-a-and-b-be-two-sequence-such-that-A-lim-n-a-n-B-lim-n-b-n-exist-and-are-finite-lets-c-be-a-new-sequence-c-n-p-n-a-n-q-n-b-n-p-q-N-0-1-p-n-q-n-1-d-N-d-N-d-d-

Question Number 4507 by 123456 last updated on 03/Feb/16 $$\mathrm{lets}\:{a}\:\mathrm{and}\:{b}\:\mathrm{be}\:\mathrm{two}\:\mathrm{sequence}\:\mathrm{such}\:\mathrm{that} \\ $$$${A}=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{a}_{{n}} \\ $$$${B}=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{b}_{{n}} \\ $$$$\mathrm{exist}\:\mathrm{and}\:\mathrm{are}\:\mathrm{finite},\:\mathrm{lets} \\ $$$${c}\:\mathrm{be}\:\mathrm{a}\:\mathrm{new}\:\mathrm{sequence} \\ $$$${c}_{{n}} ={p}\left({n}\right){a}_{\sigma\left({n}\right)} +{q}\left({n}\right){b}_{\mu\left({n}\right)} \\…

If-a-2-b-2-c-2-1-a-3-1-b-3-1-c-3-a-3-b-3-c-3-than-prove-that-a-b-c-Successive-Proportional-

Question Number 70040 by Shamim last updated on 30/Sep/19 $$\mathrm{If},\:\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:\mathrm{than} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{Successive}\:\mathrm{Proportional}. \\ $$ Commented by…