Given 3sinP+4cosQ=6 4sinQ+3cosP=1 Squaring and adding (i)&(ii) we get 9sin2P+16cos2Q+24sinPcosQ +16sin2Q+9cos2P+24sinQcosP =36+1=37 ⇒9(sin2p+cos2P)+16(sin2Q+cos2q) +24(sinPcosQ+cosPsinQ)=37 ⇒9+16+24sin(P+Q)=37 [ As sin2θ+cos2θ=1 and sinAcosB+cosAsinB=sin(A+B)] ⇒sin(P+Q)=