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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)) ?

provethatcot5x=110tan2x+5tan4xtan5x10tan3x+5tanx?

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))

tan(2x+3x)=tan2x+tan3x1tan2x.tan3xtan5x=2tanx1tan2x+3tanxtan3x13tan2x1(2tanx1tan2x)(3tanxtan3x13tan2x)=2tanx(13tan2x)+(1tan2x)(3tanxtan3x)(1tan2x)(13tan2x)2tanx(3tanxtan3x)=2tanx6tan3x+3tanx3tan3xtan3x+tan5x1tan2x3tan2x+3tan4x6tan2x+2tan4x=tan5x10tan3x+5tanx110tan2x+5tan4x

Commented by mohammad17 last updated on 24/Nov/20

thank you sir

thankyousir

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

cot5x=1tan(3x+2x)=1tan3xtan2xtan3x+tan2x=1(3tanxtan3x)(2tanx)(13tan2x)(1tan2x)3tanxtan3x13tan2x+2tanx1tan2xlettanx=a(13a2)(1a2)6a2+2a4(1a2)(3aa3)+2a6a3=1a23a2+3a46a2+2a43aa33a3+a5+2a6a3=110a2+5a45a10a3+a5nowputa=tanx

Commented by mohammad17 last updated on 24/Nov/20

thank you sir

thankyousir

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(nx)=s1s3+s5s7+...1s2+s4s6+..tan5x=s1s3+s51s2+s4s1=5C1tanx=5tanxs3=5C3tan3x=5×42tan3x=10tan3xs5=5C5tan5x=tan5xs2=5c2tan2x=10tan2xs4=5c4tan4x=5tan4xcot5x=1tan5x=110tan2x+5tan4x5tanx10tan3x+tan5x

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

Nice sir

Nicesir

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

খুব ভাল

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

thank you...

thankyou...

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