1thanx..i'll try to solve by ur method...
eliminate θ and φ if sinθ+sinφ=a , cosθ+cosφ=b and tan(θ/2)tan(φ/2)=c
the way i tried to solve this ques...has made the solution complex...
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8 Answers
62cant think of a simple soln right away.. but seems interesting..
can ewe try to introduce complex nos?
eiφ + eiθ = a+ib b+ai (edited)
I am not sure.. but can this help? just thinking!
1actually i got another easier way!!
take cosθ=x and cosφ=y
then x+y =
also
tanθ/2tanφ/2=c
hence
(sinθ/2sinφ/2)/(cosφ/2cosθ/2)=c
multiplying both numerator and denominator with 4cosθ/2cosφ/2
we get
(sinθsinφ)/(2cos2θ/2)*(2cos2φ/2)=c
replacing 2cos2θ/2 by cosθ+1 and similarly the other
we get the expression (i)
now square the first equation and the second equation and add them
you will get the relation between
sinθsinφ and cosθcosφ
thus you can write expression (i) in terms of cosθ and cosφ
thus you have two equations and two variables!!
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are ye to maine kaha tha to Vyom :P