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log-tan-x-sinx-1-log-cos-x-tanx-




Question Number 207128 by hardmath last updated on 07/May/24
log_(tan x)   sinx  −  (1/(log_(cos x)   tanx))  =  ?
$$\mathrm{log}_{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\mathrm{x}}} \:\:\mathrm{sinx}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{log}_{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}} \:\:\mathrm{tanx}}\:\:=\:\:? \\ $$
Answered by Frix last updated on 07/May/24
=((ln s)/(ln t))−((ln c)/(ln t))=((ln s −ln c)/(ln (s/c)))  If c>0∧s>0 this is equal to 1, else it′s getting  complicated.
$$=\frac{\mathrm{ln}\:{s}}{\mathrm{ln}\:{t}}−\frac{\mathrm{ln}\:{c}}{\mathrm{ln}\:{t}}=\frac{\mathrm{ln}\:{s}\:−\mathrm{ln}\:{c}}{\mathrm{ln}\:\frac{{s}}{{c}}} \\ $$$$\mathrm{If}\:{c}>\mathrm{0}\wedge{s}>\mathrm{0}\:\mathrm{this}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1},\:\mathrm{else}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\ $$$$\mathrm{complicated}. \\ $$
Commented by hardmath last updated on 07/May/24
yes dear professor >0
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor}\:>\mathrm{0} \\ $$
Commented by Frix last updated on 07/May/24
c>0∧s>0 ⇒ ln (s/c) =ln s −ln c
$${c}>\mathrm{0}\wedge{s}>\mathrm{0}\:\Rightarrow\:\mathrm{ln}\:\frac{{s}}{{c}}\:=\mathrm{ln}\:{s}\:−\mathrm{ln}\:{c} \\ $$
Commented by hardmath last updated on 07/May/24
Answer = 1 .?
$$\mathrm{Answer}\:=\:\mathrm{1}\:.? \\ $$
Commented by Frix last updated on 07/May/24
Yes for cos x >0 ∧sin x >0 ∧cos x ≠sin x
$$\mathrm{Yes}\:\mathrm{for}\:\mathrm{cos}\:{x}\:>\mathrm{0}\:\wedge\mathrm{sin}\:{x}\:>\mathrm{0}\:\wedge\mathrm{cos}\:{x}\:\neq\mathrm{sin}\:{x} \\ $$

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