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Question Number 22701 by selestian last updated on 22/Oct/17

	Answered by ajfour last updated on 22/Oct/17

	$e^{(\frac{\sin^{2} x}{1 - \sin^{2} x}) \ln 2} = e^{\ln 2^{\tan^{2} x}} = 2^{\tan^{2} x}$ $r o o t s o f x^{2} - 9 x + 8 = 0 a r e 1, 8$ $a n d 2^{\tan^{2} x} a l s o s a t i s f i e s t h e e q n .$ $S o 2^{\tan^{2} x} = 8, 1$ $\Rightarrow \tan^{2} x = 3, o r 0$ $A n d i f 0 < x < \frac{π}{2}, t h e n$ $\tan x = \sqrt{3}$ $\frac{\cos x}{\cos x + \sin x} = \frac{1}{1 + \tan x} = \frac{1}{1 + \sqrt{3}} .$
	Answered by ajfour last updated on 22/Oct/17

	$x^{2} - 3 x + 2 m = 0$ $l e t i t s r o o t s b e 2 α, β$ $x^{2} - x + m = 0$ $l e t i t s r o o t s b e α, γ$ $T h e n 2 α + β = 3, 2 α β = 2 m$ $A l s o α + β = 1, α β = m$ $(2 α + β) - (α + β) = 3 - 1$ $\Rightarrow α = 2, β = - 1 a n d m = - 2 .$
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