Question

Prove that:
(i) tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
(ii) $\tan \frac{\pi }{12}+\tan \frac{\pi }{6}+\tan \frac{\pi }{12}\tan \frac{\pi }{6}=1$
(iii) tan 36° + tan 9° + tan 36° tan 9° = 1
(iv) tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{know}\mathrm{that}8x=6x+2x \\ \mathrm{Therefore}, \\ \tan \left(8x\right)=\tan \left(6x+2x\right) \\ \Rightarrow \tan \left(8x\right)=\frac{\tan 6x+\tan 2x}{1-\tan 6x\tan 2x} \\ \Rightarrow \tan 8x-\tan 8x\tan 6x\tan 2x=\tan 6x+\tan 2x \\ \Rightarrow \tan 8x-\tan 6x-\tan 2x=\tan 8x\tan 6x\tan 2x \\ \text{Hence proved.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{\pi }{12}=15°,\frac{\pi }{6}=30° \\ \mathrm{We}\mathrm{know}\mathrm{that}45°=15°+30° \\ \mathrm{Therefore}, \\ \tan \left(45°\right)=\tan \left(15°+30°\right) \\ \Rightarrow 1=\frac{\tan 15°+\tan 30°}{1-\tan 15°\tan 30°} \\ \Rightarrow 1-\tan 15°\tan 30°=\tan 15°+\tan 30° \\ \Rightarrow 1=\tan 15°+\tan 30°+\tan 15°\tan 30° \\ \Rightarrow \tan 15°+\tan 30°+\tan 15°\tan 30°=1 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{know}\mathrm{that}36°+9°=45° \\ \mathrm{Therefore}, \\ \tan \left(36°+9°\right)=\tan 45° \\ \Rightarrow \frac{\tan 36°+\tan 9°}{1-\tan 36°\tan 9°}=1 \\ \Rightarrow \tan 36°+\tan 9°=1-\tan 36°\tan 9° \\ \Rightarrow \tan 36°+\tan 9°+\tan 36°\tan 9°=1 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{know}\mathrm{that}13x=9x+4x \\ \mathrm{Therefore}, \\ \tan \left(13x\right)=\tan \left(9x+4x\right) \\ \Rightarrow \tan 13x=\frac{\tan 9x+\tan 4x}{1-\tan 9x\tan 4x} \\ \Rightarrow \tan 13x-\tan 13x\tan 9x\tan 4x=\tan 9x+\tan 4x \\ \Rightarrow \tan 13x-\tan 9x-\tan 4x=\tan 13x\tan 9x\tan 4x \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$