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Question-86231




Question Number 86231 by liki last updated on 27/Mar/20
Commented by liki last updated on 27/Mar/20
..plz i need help qn no. 5(a) & c
..plzineedhelpqnno.5(a)&c
Answered by TANMAY PANACEA. last updated on 27/Mar/20
((tan145−tan125)/(1+tan145tan125))  =tan(145−125)  =tan20  =cot70  =(1/(tan70))  =((1−tan^2 35)/(2tan35))  =((1−x^2 )/(2x))
tan145tan1251+tan145tan125=tan(145125)=tan20=cot70=1tan70=1tan2352tan35=1x22x
Commented by liki last updated on 27/Mar/20
....Thank you so much
.Thankyousomuch
Answered by TANMAY PANACEA. last updated on 27/Mar/20
cosx+cosy=cosa  sinx+siny=sina  a=((7π)/(12))=105^o   from 1)  2cos((x+y)/2)cos((x−y)/2)=cosa  from 2) 2sin((x+y)/2)cos((x−y)/2)=sina  tan(((x+y)/2))=tana  x+y=2a=((7π)/6)=210^o   4cos^2 ((x+y)/2)cos^2 ((x−y)/2)+4sin^2 ((x+y)/2)cos^2 ((x−y)/2)=1  4cos^2 ((x−y)/2)(cos^2 ((x+y)/2)+sin^2 ((x+y)/2))=1  cos((x−y)/2)=±(1/2)  cos(((x−y)/2))=(1/2)=cos(π/3)  x−y=120^o   x+y=210^o   2x=330^o      x=165^o   y=45^o   now if  cos(((x−y)/2))=((−1)/2)=cos120^o   x−y=240^o   but look  x+y=210^o   but x−y=240^o   so cos(((x−y)/2))≠((−1)/2)  pls check...
cosx+cosy=cosasinx+siny=sinaa=7π12=105ofrom1)2cosx+y2cosxy2=cosafrom2)2sinx+y2cosxy2=sinatan(x+y2)=tanax+y=2a=7π6=210o4cos2x+y2cos2xy2+4sin2x+y2cos2xy2=14cos2xy2(cos2x+y2+sin2x+y2)=1cosxy2=±12cos(xy2)=12=cosπ3xy=120ox+y=210o2x=330ox=165oy=45onowifcos(xy2)=12=cos120oxy=240obutlookx+y=210obutxy=240osocos(xy2)12plscheck
Commented by liki last updated on 27/Mar/20
...Be blessed sir, i understood well!
Beblessedsir,iunderstoodwell!

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