Question Number 132081 by liberty last updated on 11/Feb/21
$$\mathrm{Prove}\:\mathrm{tan}\:\left(\mathrm{A}+\mathrm{B}\right)−\mathrm{tan}\:\mathrm{A}=\:\frac{\mathrm{sin}\:\mathrm{B}}{\mathrm{cos}\:\mathrm{A}\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)} \\ $$
Answered by rs4089 last updated on 11/Feb/21
$${tan}\left({A}+{B}\right)−{tanA} \\ $$$$\frac{{sin}\left({A}+{B}\right)}{{cos}\left({A}+{B}\right)}−\frac{{sinA}}{{cosA}} \\ $$$$\frac{{sin}\left({A}+{B}\right).{cosA}−{sinA}.{cos}\left({A}+{B}\right)}{{cos}\left({A}+{B}\right).{cosA}} \\ $$$$\frac{{sin}\left\{\left({A}+{B}\right)−\left({A}\right)\right\}}{{cos}\left({A}+{B}\right).{cosA}} \\ $$$$\frac{{sinB}}{{cos}\left({A}+{B}\right).{cosA}} \\ $$