Question Number 24971 by Rishabh#1 last updated on 30/Nov/17
$$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:\mathrm{3}{x}}+\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{9}{x}}+\frac{\mathrm{cos}\:\mathrm{9}{x}}{\mathrm{cos}\:\mathrm{27}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:\mathrm{27}{x}−\mathrm{tan}\:{x}\right). \\ $$
Answered by math solver last updated on 09/Feb/18
$$\mathrm{assuming}\:\mathrm{3rd}\:\mathrm{term}\:\mathrm{to}\:\mathrm{be}\: \\ $$$$\:\:\Rightarrow\:\frac{\mathrm{sin9}{x}}{{cos}\mathrm{27}{x}}. \\ $$$${we}\:{can}\:{write}\: \\ $$$$\:\frac{{sinx}}{{cos}\mathrm{3}{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\:\frac{\mathrm{2}{sinx}.{cosx}}{{cos}\mathrm{3}{x}.{cosx}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{sin2}{x}}{{cos}\mathrm{3}{x}.{cosx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{{sin}\left(\mathrm{3}{x}−{x}\right)}{{cos}\mathrm{3}{x}.{cosx}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{tan3}{x}\:−\:{tanx}\right)…….\left(\mathrm{1}\right) \\ $$$${similarly}\:, \\ $$$$\frac{\mathrm{sin3}{x}}{{cos}\mathrm{9}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left({tan}\mathrm{9}{x}−{tan}\mathrm{3}{x}\right)……..\left(\mathrm{2}\right) \\ $$$${and}, \\ $$$$\:\frac{{sin}\mathrm{9}{x}}{{cos}\:\mathrm{27}{x}}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:{tan}\mathrm{27}{x}−{tan}\mathrm{9}{x}\right)……..\left(\mathrm{3}\right) \\ $$$$\mathrm{add}\:\mathrm{all}\:\mathrm{three}\:\mathrm{eq}.\left(\mathrm{s}\right)\:, \\ $$$$\mathrm{we}\:\mathrm{get}\:, \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan27}{x}\:−\:{tan}\:{x}\right). \\ $$
Commented by Rishabh#1 last updated on 30/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$