Question

Guys PLEASE answer this question quickly.I will mark the best answer as brainliest.​

Answer 1

Question:-If sinx+siny=m,and cosx+cosy=n,, then prove that,sin(x+y)=2mn/(m^2+n^2)

Solution:-

Sinx+Siny=m

Sin^2x+Sin^2y+2SinxSiny=m^2......i)

cosx+cosy=n

cos^2x+cos^2y+2cosxcosu=n^2......ii)

multiplying both equation we get

(Sinx+Siny)(cosx+cosy)=mn

Sinxcosx+Sinxcosy+Sinycosx+Sinycosy=mn

2(Sinxcosx+Sinxcosy+Sinycosx+Sinycosy)=2mn.iii)

adding equation first and second and dividing the sum equation III) ,we get

(m^2+n^2)/2mn=(sin^2x+Sin^2y+2Sinxcosx+cos^2x+cos^2y+cos^2y+2cosxcosy)/2(sinxcosx+sinxcosy+sinycosx+sinycosy)

=>(1+1+2sinxcosx+2cosxcosy)/2(sinxcosx+sinxcosy+sinycosx+sinycosy)

=>(sinxcosx+cosxcosy)/{sinx(cosx+cosy)+Siny(cosx+cosy)

=>sinx(cosx+cosy)/(sinx+siny)(cosx+cosy)

=>1/(sinxcosy+cosysinx)

2mn/(m^2+n^2)=sinxcosy+cosxsiny

=sin(x+y)=RHS

{hope it helps}