Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 123232 by mohammad17 last updated on 24/Nov/20

prove that cot5x=((1−10tan^2 x+5tan^4 x)/(tan^5 x−10tan^3 x+5tanx)) ?

$${prove}\:{that}\:{cot}\mathrm{5}{x}=\frac{\mathrm{1}−\mathrm{10}{tan}^{\mathrm{2}} {x}+\mathrm{5}{tan}^{\mathrm{4}} {x}}{{tan}^{\mathrm{5}} {x}−\mathrm{10}{tan}^{\mathrm{3}} {x}+\mathrm{5}{tanx}}\:? \\ $$

Answered by $@y@m last updated on 24/Nov/20

tan (2x+3x)=((tan 2x+tan 3x)/(1−tan 2x.tan 3x))  ⇒ tan5x=((((2tan x)/(1−tan^2 x))+((3tan x−tan^3 x)/(1−3tan^2 x)))/(1−(((2tan x)/(1−tan^2 x)))(((3tan x−tan^3 x)/(1−3tan^2 x)))))  =((2tan x(1−3tan^2 x)+(1−tan^2 x)(3tan x−tan^3 x))/((1−tan^2 x)(1−3tan^2 x)−2tan x(3tan x−tan^3 x)))  =((2tan x−6tan^3 x+3tan x−3tan^3 x−tan^3 x+tan^5 x)/(1−tan^2 x−3tan^2 x+3tan^4 x−6tan^2  x+2tan^4 x))  =((tan^5 x−10tan^3 x+5tan x)/(1−10tan^2 x+5tan^4 x))

$$\mathrm{tan}\:\left(\mathrm{2}{x}+\mathrm{3}{x}\right)=\frac{\mathrm{tan}\:\mathrm{2}{x}+\mathrm{tan}\:\mathrm{3}{x}}{\mathrm{1}−\mathrm{tan}\:\mathrm{2}{x}.\mathrm{tan}\:\mathrm{3}{x}} \\ $$$$\Rightarrow\:\mathrm{tan5}{x}=\frac{\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}+\frac{\mathrm{3tan}\:{x}−\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}}{\mathrm{1}−\left(\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\right)\left(\frac{\mathrm{3tan}\:{x}−\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}\right)} \\ $$$$=\frac{\mathrm{2tan}\:{x}\left(\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}\right)+\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right)\left(\mathrm{3tan}\:{x}−\mathrm{tan}\:^{\mathrm{3}} {x}\right)}{\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right)\left(\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}\right)−\mathrm{2tan}\:{x}\left(\mathrm{3tan}\:{x}−\mathrm{tan}\:^{\mathrm{3}} {x}\right)} \\ $$$$=\frac{\mathrm{2tan}\:{x}−\mathrm{6tan}^{\mathrm{3}} {x}+\mathrm{3tan}\:{x}−\mathrm{3tan}^{\mathrm{3}} {x}−\mathrm{tan}^{\mathrm{3}} {x}+\mathrm{tan}^{\mathrm{5}} {x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}−\mathrm{3tan}\:^{\mathrm{2}} {x}+\mathrm{3tan}\:^{\mathrm{4}} {x}−\mathrm{6tan}^{\mathrm{2}} \:{x}+\mathrm{2tan}^{\mathrm{4}} {x}} \\ $$$$=\frac{\mathrm{tan}\:^{\mathrm{5}} {x}−\mathrm{10tan}\:^{\mathrm{3}} {x}+\mathrm{5tan}\:{x}}{\mathrm{1}−\mathrm{10tan}\:^{\mathrm{2}} {x}+\mathrm{5tan}\:^{\mathrm{4}} {x}} \\ $$

Commented by mohammad17 last updated on 24/Nov/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by som(math1967) last updated on 24/Nov/20

cot5x  =(1/(tan(3x+2x)))=((1−tan3xtan2x)/(tan 3x+tan 2x))  =((1−(((3tanx−tan^3 x)(2tanx))/((1−3tan^2 x)(1−tan^2 x))))/(((3tanx−tan^3 x)/(1−3tan^2 x))+((2tanx)/(1−tan^2 x))))  let tanx=a  (((1−3a^2 )(1−a^2 )−6a^2 +2a^4 )/((1−a^2 )(3a−a^3 )+2a−6a^3 ))  =((1−a^2 −3a^2 +3a^4 −6a^2 +2a^4 )/(3a−a^3 −3a^3 +a^5 +2a−6a^3 ))  =((1−10a^2 +5a^4 )/(5a−10a^3 +a^5 ))  now put a=tanx

$$\mathrm{cot5x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{tan}\left(\mathrm{3x}+\mathrm{2x}\right)}=\frac{\mathrm{1}−\mathrm{tan3xtan2x}}{\mathrm{tan}\:\mathrm{3x}+\mathrm{tan}\:\mathrm{2x}} \\ $$$$=\frac{\mathrm{1}−\frac{\left(\mathrm{3tanx}−\mathrm{tan}^{\mathrm{3}} \mathrm{x}\right)\left(\mathrm{2tanx}\right)}{\left(\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}}{\frac{\mathrm{3tanx}−\mathrm{tan}^{\mathrm{3}} \mathrm{x}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{2tanx}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{x}}} \\ $$$$\mathrm{let}\:\mathrm{tanx}=\mathrm{a} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{3a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)−\mathrm{6a}^{\mathrm{2}} +\mathrm{2a}^{\mathrm{4}} }{\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{3a}−\mathrm{a}^{\mathrm{3}} \right)+\mathrm{2a}−\mathrm{6a}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{a}^{\mathrm{2}} −\mathrm{3a}^{\mathrm{2}} +\mathrm{3a}^{\mathrm{4}} −\mathrm{6a}^{\mathrm{2}} +\mathrm{2a}^{\mathrm{4}} }{\mathrm{3a}−\mathrm{a}^{\mathrm{3}} −\mathrm{3a}^{\mathrm{3}} +\mathrm{a}^{\mathrm{5}} +\mathrm{2a}−\mathrm{6a}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{10a}^{\mathrm{2}} +\mathrm{5a}^{\mathrm{4}} }{\mathrm{5a}−\mathrm{10a}^{\mathrm{3}} +\mathrm{a}^{\mathrm{5}} } \\ $$$$\mathrm{now}\:\mathrm{put}\:\mathrm{a}=\mathrm{tanx} \\ $$

Commented by mohammad17 last updated on 24/Nov/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by TANMAY PANACEA last updated on 24/Nov/20

tan(nx)=((s_1 −s_3 +s_5 −s_7 +...)/(1−s_2 +s_4 −s_6 +..))  tan5x=((s_1 −s_3 +s_5 )/(1−s_2 +s_4 ))  s_1 =5C_1 tanx=5tanx  s_3 =5C_3 tan^3 x=((5×4)/2)tan^3 x=10tan^3 x  s_5 =5C_5 tan^5 x=tan^5 x  s_2 =5c_2 tan^2 x=10tan^2 x  s_4 =5c_4 tan^4 x=5tan^4 x  cot5x=(1/(tan5x))=((1−10tan^2 x+5tan^4 x)/(5tanx−10tan^3 x+tan^5 x))

$${tan}\left({nx}\right)=\frac{{s}_{\mathrm{1}} −{s}_{\mathrm{3}} +{s}_{\mathrm{5}} −{s}_{\mathrm{7}} +...}{\mathrm{1}−{s}_{\mathrm{2}} +{s}_{\mathrm{4}} −{s}_{\mathrm{6}} +..} \\ $$$${tan}\mathrm{5}{x}=\frac{{s}_{\mathrm{1}} −{s}_{\mathrm{3}} +{s}_{\mathrm{5}} }{\mathrm{1}−{s}_{\mathrm{2}} +{s}_{\mathrm{4}} } \\ $$$${s}_{\mathrm{1}} =\mathrm{5}{C}_{\mathrm{1}} {tanx}=\mathrm{5}{tanx} \\ $$$${s}_{\mathrm{3}} =\mathrm{5}{C}_{\mathrm{3}} {tan}^{\mathrm{3}} {x}=\frac{\mathrm{5}×\mathrm{4}}{\mathrm{2}}{tan}^{\mathrm{3}} {x}=\mathrm{10}{tan}^{\mathrm{3}} {x} \\ $$$${s}_{\mathrm{5}} =\mathrm{5}{C}_{\mathrm{5}} {tan}^{\mathrm{5}} {x}={tan}^{\mathrm{5}} {x} \\ $$$${s}_{\mathrm{2}} =\mathrm{5}{c}_{\mathrm{2}} {tan}^{\mathrm{2}} {x}=\mathrm{10}{tan}^{\mathrm{2}} {x} \\ $$$${s}_{\mathrm{4}} =\mathrm{5}{c}_{\mathrm{4}} {tan}^{\mathrm{4}} {x}=\mathrm{5}{tan}^{\mathrm{4}} {x} \\ $$$${cot}\mathrm{5}{x}=\frac{\mathrm{1}}{{tan}\mathrm{5}{x}}=\frac{\mathrm{1}−\mathrm{10}{tan}^{\mathrm{2}} {x}+\mathrm{5}{tan}^{\mathrm{4}} {x}}{\mathrm{5}{tanx}−\mathrm{10}{tan}^{\mathrm{3}} {x}+{tan}^{\mathrm{5}} {x}} \\ $$

Commented by som(math1967) last updated on 25/Nov/20

Nice sir

$$\mathrm{Nice}\:\mathrm{sir}\: \\ $$

Commented by som(math1967) last updated on 25/Nov/20

খুব ভাল

Commented by TANMAY PANACEA last updated on 25/Nov/20

thank you...

$${thank}\:{you}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com