y^′ =((cosy−siny−1)/(cosx−sinx+1))


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Question Number 121827 by bounhome last updated on 12/Nov/20

$y^{^{'}} = \frac{c o s y - s i n y - 1}{c o s x - s i n x + 1}$
	Answered by TANMAY PANACEA last updated on 12/Nov/20
	$(dy/dx)=((cosy−siny−1)/(cosx−sinx+1)) (dy/(cosy−siny−1))=(dx/(cosx−sinx+1)) b=tan(y/2) a=tan(x/2) db=sec^2 ((y/2))×(1/2)×dy→dy=((2db)/(1+b^2 )) & dx=((2da)/(1+a^2 )) ∫((2db)/((((1−b^2 )/(1+b^2 ))−((2b)/(1+b^2 ))−1)×(1+b^2 )))=∫((2da)/((((1−a^2 )/(1+a^2 ))−((2a)/(1+a^2 ))+1)×(1+a^2 ))) ∫((2db)/(1−b^2 −2b−1−b^2 ))=∫((2da)/(1−a^2 −2a+1+a^2 )) ∫(db/(−2b^2 −2b))=∫(da/(2−2a)) ∫(db/(b^2 +b))=∫(da/(a−1)) ∫((b+1−b)/(b(b+1)))db=∫(da/(a−1)) ∫(db/b)−∫(db/(b+1))=∫(da/(a−1)) lnb−ln(b+1)=ln(a−1)+lnC ln((b/(b+1)))=ln{C×(a−1)} (b/(b+1))=C(a−1) ((tan(y/2))/(1+tan(y/2)))=C(tan(x/2)−1)$
	$\frac{d y}{d x} = \frac{c o s y - s i n y - 1}{c o s x - s i n x + 1}$ $\frac{d y}{c o s y - s i n y - 1} = \frac{d x}{c o s x - s i n x + 1}$ $b = t a n \frac{y}{2} a = t a n \frac{x}{2}$ $d b = {s e c}^{2} (\frac{y}{2}) \times \frac{1}{2} \times d y \to d y = \frac{2 d b}{1 + b^{2}} & d x = \frac{2 d a}{1 + a^{2}}$ $\int \frac{2 d b}{(\frac{1 - b^{2}}{1 + b^{2}} - \frac{2 b}{1 + b^{2}} - 1) \times (1 + b^{2})} = \int \frac{2 d a}{(\frac{1 - a^{2}}{1 + a^{2}} - \frac{2 a}{1 + a^{2}} + 1) \times (1 + a^{2})}$ $\int \frac{2 d b}{1 - b^{2} - 2 b - 1 - b^{2}} = \int \frac{2 d a}{1 - a^{2} - 2 a + 1 + a^{2}}$ $\int \frac{d b}{- 2 b^{2} - 2 b} = \int \frac{d a}{2 - 2 a}$ $\int \frac{d b}{b^{2} + b} = \int \frac{d a}{a - 1}$ $\int \frac{b + 1 - b}{b (b + 1)} d b = \int \frac{d a}{a - 1}$ $\int \frac{d b}{b} - \int \frac{d b}{b + 1} = \int \frac{d a}{a - 1}$ $l n b - l n (b + 1) = l n (a - 1) + l n C$ $l n (\frac{b}{b + 1}) = l n {C \times (a - 1)}$ $\frac{b}{b + 1} = C (a - 1)$ $\frac{t a n \frac{y}{2}}{1 + t a n \frac{y}{2}} = C (t a n \frac{x}{2} - 1)$
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