Prove that:

Question:

(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

Solution:

(i) We know that $8 x=6 x+2 x$

Therefore,

$\tan (8 x)=\tan (6 x+2 x)$

$\Rightarrow \tan (8 x)=\frac{\tan 6 x+\tan 2 x}{1-\tan 6 x \tan 2 x}$

$\Rightarrow \tan 8 x-\tan 8 x \tan 6 x \tan 2 x=\tan 6 x+\tan 2 x$

$\Rightarrow \tan 8 x-\tan 6 x-\tan 2 x=\tan 8 x \tan 6 x \tan 2 x$

Hence proved.

(ii) $\frac{\pi}{12}=15^{\circ}, \frac{\pi}{6}=30^{\circ}$

We know that $45^{\circ}=15^{\circ}+30^{\circ}$

Therefore

$\tan \left(45^{\circ}\right)=\tan \left(15^{\circ}+30^{\circ}\right)$

$\Rightarrow 1=\frac{\tan 15^{\circ}+\tan 30^{\circ}}{1-\tan 15^{\circ} \tan 30^{\circ}}$

$\Rightarrow 1-\tan 15^{\circ} \tan 30^{\circ}=\tan 15^{\circ}+\tan 30^{\circ}$

$\Rightarrow 1=\tan 15^{\circ}+\tan 30^{\circ}+\tan 15^{\circ} \tan 30^{\circ}$

$\Rightarrow \tan 15^{\circ}+\tan 30^{\circ}+\tan 15^{\circ} \tan 30^{\circ}=1$

Hence proved.

(iii) We know that $36^{\circ}+9^{\circ}=45^{\circ}$

Therefore,

$\tan \left(36^{\circ}+9^{\circ}\right)=\tan 45^{\circ}$

$\Rightarrow \frac{\tan 36^{\circ}+\tan 9^{\circ}}{1-\tan 36^{\circ} \tan 9^{\circ}}=1$

$\Rightarrow \tan 36^{\circ}+\tan 9^{\circ}=1-\tan 36^{\circ} \tan 9^{\circ}$

$\Rightarrow \tan 36^{\circ}+\tan 9^{\circ}+\tan 36^{\circ} \tan 9^{\circ}=1$

Hence proved.

(iv) We know that $13 x=9 x+4 x$

Therefore,

$\tan (13 x)=\tan (9 x+4 x)$

$\Rightarrow \tan 13 x=\frac{\tan 9 x+\tan 4 x}{1-\tan 9 x \tan 4 x}$

$\Rightarrow \tan 13 x-\tan 13 x \tan 9 x \tan 4 x=\tan 9 x+\tan 4 x$

$\Rightarrow \tan 13 x-\tan 9 x-\tan 4 x=\tan 13 x \tan 9 x \tan 4 x$

Hence proved.