Menu Close

Category: Arithmetic

Question-68210

Question Number 68210 by peter frank last updated on 07/Sep/19 Answered by @ty@m123lastupdatedon08/Sep/19LetsinAa=sinBb=sinCc=R$sinA=aR,sinB=bR,sinC=cR(1)(2)sin(AB)sinC1+cos(AB)cosC$$=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:\left({A}+{B}\right)}{\mathrm{1}−\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:\left({A}+{B}\right)} \

Question-68191

Question Number 68191 by peter frank last updated on 06/Sep/19 Answered by mind is power last updated on 06/Sep/19 sin(ab)=sin(a)cos(b)cos(a)sin(b)sin(ab)cos(a)cos(b)=tg(a)tg(b)$$\Rightarrow\frac{{sin}\left({a}−{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}={tg}\left({b}\right)\left({k}−\mathrm{1}\right) \