Prove that tan142°30′+(√6)+(√3)−(√2) is an integer.


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Question Number 109914 by 1549442205PVT last updated on 26/Aug/20

$Prove that \tan 142 ° 30^{'} + \sqrt{6} + \sqrt{3} - \sqrt{2}$ $is an integer .$
	Answered by Dwaipayan Shikari last updated on 26/Aug/20

	$- t a n (37.5) ° = - t a n \frac{5 π}{24}$ $1 - c o s \frac{5 π}{12} = 2 {s i n}^{2} \frac{5 π}{24}$ $t a n \frac{5 π}{24} = \frac{s i n \frac{5 π}{24}}{c o s \frac{5 π}{24}} = \frac{2 {s i n}^{2} \frac{5 π}{24}}{2 c o s \frac{5 π}{24} s i n \frac{5 π}{24}} = \frac{1 - c o s \frac{5 π}{12}}{s i n \frac{5 π}{12}} = \frac{1 - \frac{\sqrt{3} - 1}{2 \sqrt{2}}}{\frac{\sqrt{3} + 1}{2 \sqrt{2}}} = \frac{2 \sqrt{2} - \sqrt{3} + 1}{\sqrt{3} + 1}$ $= \frac{2 \sqrt{2} - \sqrt{3} + 1}{2} (\sqrt{3} - 1) = \frac{2 \sqrt{6} - 2 \sqrt{2} - (4 - 2 \sqrt{3})}{2} = \sqrt{6} - \sqrt{2} - 2 + \sqrt{3}$ $- t a n \frac{5 π}{24} = - \sqrt{6} + \sqrt{2} + 2 - \sqrt{3}$ $t a n 142.5 ° + \sqrt{6} + \sqrt{3} - \sqrt{2} = 2$
	Commented by 1549442205PVT last updated on 27/Aug/20

	$Thank Sir .$
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