Question
Prove that: $\tan13\text{x}-\tan9\text{x}-\tan4\text{x}=\tan13\text{x}\tan9\text{x}\tan4\text{x}$
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Answer
We have, $13\text{x}=9\text{x}+4\text{x}$ $\Rightarrow\tan13\text{x}=\tan(9\text{x}+4\text{x)}$ $\Rightarrow\tan13\text{x}=\frac{\tan9\text{x}+\tan4\text{x}}{1-\tan9\text{x}\tan4\text{x}}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$ $\Rightarrow\tan13\text{x}(1-\tan9\text{x}\tan4\text{x)}=\tan9\text{x}+\tan4\text{x}$ $\Rightarrow\tan13\text{x}-\tan13\text{x} \tan9\text{x} \tan4\text{x}=\tan9\text{x}+\tan4\text{x}$ $\Rightarrow\tan13\text{x}-\tan9\text{x}-\tan4\text{x}=\tan13\text{x}\tan9\text{x}\tan4\text{x}$ Hence proved.
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