If tan θ=((cos 9°+sin 9°)/(cos 9°−sin 9°)) then θ = __


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Question Number 49459 by Pk1167156@gmail.com last updated on 07/Dec/18

$If \tan θ = \frac{\cos 9 ° + \sin 9 °}{\cos 9 ° - \sin 9 °} then θ =___$
	Answered by math1967 last updated on 07/Dec/18

	$t a n θ = \frac{c o s 9 + c o s (90 - 9)}{c o s 9 - c o s (90 - 9)}$ $t a n θ = \frac{c o s 9 + c o s 81}{c o s 9 - c o s 81}$ $t a n θ = \frac{2 c o s 45 . c o s 36}{2 s i n 45 . s i n 36}$ $t a n θ = c o t 36 [∵ c o s 45 = s i n 45]$ $t a n θ = c o t 36$ $t a n θ = t a n (90 - 36)$ $t a n θ = t a n 54 ∴ θ = 54 a n s$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

	$t a n θ = \frac{1 + t a n 9^{o}}{1 - t a n 9^{o}} = t a n (45^{o} + 9^{o}) = t a n 54^{o}$
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