Question
Prove that: $\tan8\text{x}-\tan6\text{x}-\tan2\text{x}=\tan8\text{x}\tan6\text{x}\tan2\text{x}$
✓
Answer
We have,
$8\text{x}=6\text{x}+2\text{x}$
$\Rightarrow\tan8\text{x}=\tan(6\text{x}+2\text{x)}$
$\Rightarrow\tan8\text{x}=\frac{\tan6\text{x}+\tan2\text{x}}{1-\tan6\text{x}\tan2\text{x}}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$\Rightarrow\tan8\text{x}(1-\tan6\text{x}\tan2\text{x)}=\tan6\text{x}+\tan2\text{x}$
$\Rightarrow\tan8\text{x}-\tan8\text{x} \tan6\text{x} \tan2\text{x}=\tan6\text{x}+\tan2\text{x}$
$\Rightarrow\tan8\text{x}-\tan6\text{x}-\tan2\text{x}=\tan8\text{x}\tan6\text{x}\tan2\text{x}$
Hence proved.
$8\text{x}=6\text{x}+2\text{x}$
$\Rightarrow\tan8\text{x}=\tan(6\text{x}+2\text{x)}$
$\Rightarrow\tan8\text{x}=\frac{\tan6\text{x}+\tan2\text{x}}{1-\tan6\text{x}\tan2\text{x}}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$\Rightarrow\tan8\text{x}(1-\tan6\text{x}\tan2\text{x)}=\tan6\text{x}+\tan2\text{x}$
$\Rightarrow\tan8\text{x}-\tan8\text{x} \tan6\text{x} \tan2\text{x}=\tan6\text{x}+\tan2\text{x}$
$\Rightarrow\tan8\text{x}-\tan6\text{x}-\tan2\text{x}=\tan8\text{x}\tan6\text{x}\tan2\text{x}$
Hence proved.
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