Mathematic Induction Question (1 Viewer)

KloppsAndRobbers · Oct 17, 2023

Been trying for ages - atp I just want to see it solved. Please help me lol.

member 6003 · Oct 17, 2023

Assume for n=k

$\tan{\theta}\tan{2\theta}+...+\tan{k\theta}\tan{(k+1)\theta}=\tan{(k+1)\theta}\cot{\theta}-(k+1)$

To prove for n=k+1

That is
$\tan{\theta}\tan{2\theta}+...+\tan{(k+1)\theta}\tan{(k+2)\theta}=\tan{(k+2)\theta}\cot{\theta}-(k+1)$

$LHS = \tan{\theta}\tan{2\theta}+...+\tan{k\theta}\tan{(k+1)\theta}+\tan{(k+1)\theta}\tan{(k+2)\theta}$

using assumption,

$= \tan{(k+1)\theta}\cot{\theta}-(k+1)+ \tan{(k+1)\theta}\tan{(k+2)\theta}$

$= -(k+2) + 1 + \tan{(k+1)\theta}\cot{\theta} + \tan{(k+1)\theta}\tan{(k+2)\theta}$

$= -(k+2) + \cot{\theta}[ \tan{\theta} + \tan{(k+1)\theta} + \tan{(k+1)\theta}\tan{(k+2)\theta}\tan{\theta} ]$

using i) for $\alpha = \theta , \beta = (k+1)\theta$

$= -(k+2) + \cot{\theta}[(\tan{(k+2)\theta})(1-\tan{\theta}\tan{(k+1)\theta}) + \tan{(k+1)\theta}\tan{(k+2)\theta}\tan{\theta}]$

$= -(k+2) + \cot{\theta}[\tan{(k+2)\theta} - \tan{\theta}\tan{(k+1)\theta}\tan{(k+2)\theta} + \tan{(k+1)\theta}\tan{(k+2)\theta}\tan{\theta}]$

$= \tan{(k+2)\theta}\cot{\theta} - (k+2)$

$= RHS$

therefore result follows by induction

sorry for bracket mistakes, this question you just want try to match very closely to the RHS or your not gonna get it because of the +1 term

KloppsAndRobbers · Oct 17, 2023

Thank you so much

Mathematic Induction Question (1 Viewer)

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