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m-2-sin-2m-1-cos-2m-1-if-1-tan-3-4-2-tan-4-3-3-tan-2m-m-2-1-4-tan-2m-m-2-1-




Question Number 18184 by Tinkutara last updated on 16/Jul/17
(m+2)sinθ + (2m−1)cosθ = 2m+1, if  (1) tanθ = (3/4)  (2) tanθ = (4/3)  (3) tanθ = ((2m)/(m^2  − 1))  (4) tanθ = ((2m)/(m^2  + 1))
(m+2)sinθ+(2m1)cosθ=2m+1,if(1)tanθ=34(2)tanθ=43(3)tanθ=2mm21(4)tanθ=2mm2+1
Commented by prakash jain last updated on 16/Jul/17
tan (θ/2)=u  sin θ=((2u)/(1+u^2 )), cos θ=((1−u^2 )/(1+u^2 ))  (m+2)2u+(2m−1)(1−u^2 )=(2m+1)(1+u^2 )  4mu^2 +2(m+2)u+2=0  2mu^2 −(m+2)u+1=0  (2u−1)(mu−1)=0  u=(1/2),(1/m)  tan θ=((2tan θ/2)/(1−tan^2 θ/2))  tan θ=(4/3),((2m)/(m^2 −1))
tanθ2=usinθ=2u1+u2,cosθ=1u21+u2(m+2)2u+(2m1)(1u2)=(2m+1)(1+u2)4mu2+2(m+2)u+2=02mu2(m+2)u+1=0(2u1)(mu1)=0u=12,1mtanθ=2tanθ/21tan2θ/2tanθ=43,2mm21
Commented by Tinkutara last updated on 17/Jul/17
Thanks Sir!
ThanksSir!
Answered by ajfour last updated on 16/Jul/17
(√((m+2)^2 +(2m−1)^2 ))sin (θ+tan^(−1) ((2m−1)/(m+2)))=2m+1  θ=−tan^(−1) ((2m−1)/(m+2))+sin^(−1) ((2m+1)/( (√((m−2)^2 +(2m+1)^2 ))))    =−tan^(−1) ((2m−1)/(m+2))+tan^(−1) ((2m+1)/(m−2))  tan θ=((((2m+1)/(m−2))−((2m−1)/(m+2)))/(1+((2m+1)/(m−2)).((2m−1)/(m+2))))    = (((2m+1)(m+2)−(2m−1)(m−2))/(m^2 −4+4m^2 −1))   tan θ=((10m)/(5(m^2 −1))) =((2m)/(m^2 −1)) .      [ option  (3) ,(2) ]     for  m=2 ,   tan θ=(4/3).
(m+2)2+(2m1)2sin(θ+tan12m1m+2)=2m+1θ=tan12m1m+2+sin12m+1(m2)2+(2m+1)2=tan12m1m+2+tan12m+1m2tanθ=2m+1m22m1m+21+2m+1m2.2m1m+2=(2m+1)(m+2)(2m1)(m2)m24+4m21tanθ=10m5(m21)=2mm21.[option(3),(2)]form=2,tanθ=43.
Commented by ajfour last updated on 16/Jul/17
you can view solution of Q.18111  (chord AB at ∠=φ with diameter  cut in the ratio p:q )
youcanviewsolutionofQ.18111(chordABat=ϕwithdiametercutintheratiop:q)
Commented by Tinkutara last updated on 16/Jul/17
I viewed it. Nice method! And thanks  for this solution!
Iviewedit.Nicemethod!Andthanksforthissolution!

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