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The discussion revolves around the application of the trigonometric identity for cotangent, specifically $\cot 2\theta = \frac{\cot^{2} \theta - 1}{2\cot \theta}$. Participants calculate values based on the identity, using $\alpha = \tan^{-1} \frac{1}{2}$ to derive results for DP and CP, both equating to 6. The conversation highlights the importance of understanding trigonometric identities in problem-solving. The exchange concludes with expressions of gratitude among participants for clarifying the calculations. Overall, the thread emphasizes the practical use of trigonometric identities in mathematical computations.
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Albert said:

we have $\tan \alpha= \frac{1}{2} $
hence $\tan 2\alpha= \frac{2*\frac{1}{2}}{1-\frac{1}{4}}= \frac{4}{3} $

hence $DP = AD\tan\,2 \alpha =6$
so CP = 2
 
Albert said:

[sp]Remembering the trigonometric identity...

$\displaystyle \cot 2\ \theta = \frac{\cot^{2} \theta - 1}{2\ \cot \theta}\ (1)$

... because is $\displaystyle \alpha = \tan^{-1} \frac{1}{2}$ You have...

$\displaystyle \text{DP} = 8\ \cot 2\ \alpha = 8\ \frac{3}{4} = 6$[/sp]

Kind regards

$\chi$ $\sigma$
 
Last edited:
chisigma said:
[sp]Remembering the trigonometric identity...

$\displaystyle \cot 2\ \theta = \frac{\cot^{2} \theta - 1}{2\ \cot \theta}\ (1)$

... because is $\displaystyle \alpha = \tan^{-1} \frac{1}{2}$ You have...

$\displaystyle \text{CP} = 8\ \cot 2\ \alpha = 8\ \frac{3}{4} = 6$[/sp]

Kind regards

$\chi$ $\sigma$

DP = 6 hence CP = 2
 
thanks, kaliprasad and chisigma
you got it !
CP=2
 
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