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\[Tan3\Theta=(3\tan \Theta-\tan ^{3}\Theta) \div(1-3\tan ^{2}\Theta)\]
\[\tan(x+y)=\frac{tanx+tany}{1-tanxtany}\] x=y=theta \[\tan(2 \theta)=\frac{2\tan \theta}{1-\tan^2 \theta}\] 7ot x=2 theta y=theta \[\huge{\tan(3 \theta)=\frac{\frac{2\tan \theta}{(1-\tan^2 \theta+\tan \theta)}}{\frac{1-2\tan^2 \theta}{1-\tan^2 \theta}}}\]simpligy lol
simplify*
allah ,sho had ya lanaaa,aked el super lel super :DDD. tysmm @lalaly
hehe anytime la sadeeqi :)
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