show that− sin 10−(√3)sec10=4.


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Question Number 34066 by 5a3k last updated on 30/Apr/18

$show that -$ $\sin 10 - \sqrt{3} \sec 10 = 4 .$
	Answered by math1967 last updated on 30/Apr/18

	$c o r r e c t q . i s \frac{1}{s i n 10 °} - \sqrt{3} s e c 10 °$ $\frac{c o s 10 ° - \sqrt{3} s i n 10}{s i n 10 c o s 10}$ $\frac{2 (\frac{1}{2} c o s 10 - \frac{\sqrt{3}}{2} s i n 10)}{s i n 10 c o s 10}$ $\frac{2 (s i n 30 c o s 10 - c o s 30 s i n 10)}{s i n 10 c o s 10}$ $\frac{2 s i n (30 - 10)}{s i n 10 c o s 10} = \frac{4 s i n 20}{2 s i n 10 c o s 10} = \frac{4 s i n 20}{s i n 20}$ $= 4$
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