if (a−b)sin(θ+φ)=(a+b)sin(θ−φ) and a tan(θ/2) − b tan(φ/2) = c then prove that the following i) sinφ = ((2bc)/(a^2 −b^2 −c^2 )) ii) sinθ = ((2ac)/(a^2 −b^2 +c^2 ))


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Question Number 144305 by gsk2684 last updated on 24/Jun/21

$if (a - b) \sin (θ + ϕ) = (a + b) \sin (θ - ϕ)$ $and a \tan \frac{θ}{2} - b \tan \frac{ϕ}{2} = c then$ $prove that the following$ $i) \sin ϕ = \frac{2 bc}{a^{2} - b^{2} - c^{2}}$ $ii) \sin θ = \frac{2 ac}{a^{2} - b^{2} + c^{2}}$
	Answered by liberty last updated on 24/Jun/21

	$\Leftrightarrow (a - b) (\sin θ \cos ϕ + \cos θ \sin ϕ) =$ $(a + b) (\sin θ \cos ϕ - \cos θ \sin ϕ)$ $\Leftrightarrow a \cos θ \sin ϕ - bsin θ \cos ϕ =$ $- acos θ \sin ϕ + bsin θ \cos ϕ$ $\Leftrightarrow 2 a \cos θ \sin ϕ = 2 b \sin θ \cos ϕ$ $\Leftrightarrow \frac{a}{b} = \frac{\sin θ \cos ϕ}{\cos θ \sin ϕ}$ $\Leftrightarrow \frac{a}{b} = \tan θ \cot ϕ$ $\Leftrightarrow a \tan ϕ = b \tan θ$ $\Leftrightarrow a (\frac{2 \tan \frac{ϕ}{2}}{1 - \tan^{2} \frac{ϕ}{2}}) = b (\frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}}) . . . (i)$ $\tan (\frac{θ}{2}) = \frac{c}{a} + \frac{b}{a} \tan (\frac{ϕ}{2}) . . . (ii)$ $let \tan (\frac{ϕ}{2}) = x$ $\Leftrightarrow a (\frac{x}{1 - x^{2}}) = b (\frac{\frac{c}{a} + \frac{b}{a} x}{1 - {(\frac{c + bx}{a})}^{2}})$ $\Leftrightarrow a (\frac{x}{1 - x^{2}}) = b (\frac{a (c + bx)}{a^{2} - c^{2} - 2 bcx - b^{2} x^{2}})$ $\Leftrightarrow \frac{x}{1 - x^{2}} = \frac{bc + b^{2} x}{a^{2} - c^{2} - 2 bcx - b^{2} x^{2}}$ $\Leftrightarrow a^{2} x - c^{2} x - {2 bcx}^{2} - b^{2} x^{3} = bc + b^{2} x - {bcx}^{2} - b^{2} x^{3}$ $\Rightarrow a^{2} x - c^{2} x - b^{2} x - {bcx}^{2} - bc = 0$ $\Rightarrow {bcx}^{2} + (b^{2} + c^{2} - a^{2}) x + bc = 0$ $\Rightarrow x = \frac{a^{2} - b^{2} - c^{2} + \sqrt{{(b^{2} + c^{2} - a^{2})}^{2} - 4 {(bc)}^{2}}}{2 bc}$ $\Rightarrow x = \frac{a^{2} - b^{2} - c^{2} + \sqrt{(b^{2} + c^{2} - a^{2} + 2 bc) (b^{2} + c^{2} - a^{2} - 2 bc)}}{2 bc}$ $\Rightarrow x = \frac{a^{2} - b^{2} - c^{2} + \sqrt{({(b + c)}^{2} - a^{2}) ({(b - c)}^{2} - a^{2})}}{2 bc}$
	Commented by gsk2684 last updated on 24/Jun/21

	$thank you very much$
	Answered by ajfour last updated on 24/Jun/21
	$((sin (θ+φ))/(sin (θ−φ)))=((a+b)/(a−b)) ⇒ ((sin (θ+φ)+sin (θ−φ))/(sin (θ+φ)−sin (θ−φ))) =((2a)/(2b))=(a/b) ⇒ ((sin θcos φ)/(cos θsin φ))=(a/b) ⇒ (((2tan (θ/2))/(1−tan^2 (θ/2))))(((1−tan^2 (φ/2))/(2tan (φ/2))))=(a/b) ((p(1−q^2 ))/(q(1−p^2 )))=(a/b) ....(I) & atan (θ/2)−btan (φ/2)=c ⇒ ap−bq=c .....(II) ((p{b^2 −(ap−c)^2 })/((ap−c)(1−p^2 )))=a b^2 p−a^2 p^3 +2acp^2 −c^2 p =−a^2 p^3 +acp^2 +a^2 p−ac ⇒ acp^2 +(b^2 −c^2 −a^2 )p+ac=0 ⇒ p+(1/p)=((a^2 +c^2 −b^2 )/(ac)) sin θ=((2p)/(1+p^2 ))=(2/((1/p)+p)) ⇒ sin𝛉=((2ac)/(a^2 −b^2 +c^2 )) similarly from (I), (II) (((c+bq)(1−q^2 ))/(q{a^2 −(c+bq)^2 }))=(1/b) ⇒ −b^2 q^3 −bcq^2 +b^2 q+bc =−b^2 q^3 −2bcq^2 +a^2 q−c^2 q ⇒ bcq^2 +(b^2 −a^2 +c^2 )q+bc=0 ⇒ q+(1/q)=((a^2 −b^2 −c^2 )/(bc)) sin φ=((2q)/(1+q^2 ))=(2/((1/q)+q)) ⇒ sin𝛗=((2bc)/(a^2 −b^2 −c^2 )) ★$
	$\frac{\sin (θ + ϕ)}{\sin (θ - ϕ)} = \frac{a + b}{a - b}$ $\Rightarrow \frac{\sin (θ + ϕ) + \sin (θ - ϕ)}{\sin (θ + ϕ) - \sin (θ - ϕ)}$ $= \frac{2 a}{2 b} = \frac{a}{b}$ $\Rightarrow \frac{\sin θ \cos ϕ}{\cos θ \sin ϕ} = \frac{a}{b}$ $\Rightarrow (\frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}}) (\frac{1 - \tan^{2} \frac{ϕ}{2}}{2 \tan \frac{ϕ}{2}}) = \frac{a}{b}$ $\frac{p (1 - q^{2})}{q (1 - p^{2})} = \frac{a}{b} . . . . (I)$ $&$ $a \tan \frac{θ}{2} - b \tan \frac{ϕ}{2} = c$ $\Rightarrow a p - b q = c . . . . . (I I)$ $\frac{p {b^{2} - {(a p - c)}^{2}}}{(a p - c) (1 - p^{2})} = a$ $b^{2} p - a^{2} p^{3} + 2 {a c p}^{2} - c^{2} p$ $= - a^{2} p^{3} + {a c p}^{2} + a^{2} p - a c$ $\Rightarrow {a c p}^{2} + (b^{2} - c^{2} - a^{2}) p + a c = 0$ $\Rightarrow p + \frac{1}{p} = \frac{a^{2} + c^{2} - b^{2}}{a c}$ $\sin θ = \frac{2 p}{1 + p^{2}} = \frac{2}{\frac{1}{p} + p}$ $\Rightarrow s i n θ = \frac{2 a c}{a^{2} - b^{2} + c^{2}}$ $s i m i l a r l y f r o m (I), (I I)$ $\frac{(c + b q) (1 - q^{2})}{q {a^{2} - {(c + b q)}^{2}}} = \frac{1}{b}$ $\Rightarrow - b^{2} q^{3} - {b c q}^{2} + b^{2} q + b c$ $= - b^{2} q^{3} - 2 {b c q}^{2} + a^{2} q - c^{2} q$ $\Rightarrow {b c q}^{2} + (b^{2} - a^{2} + c^{2}) q + b c = 0$ $\Rightarrow q + \frac{1}{q} = \frac{a^{2} - b^{2} - c^{2}}{b c}$ $\sin ϕ = \frac{2 q}{1 + q^{2}} = \frac{2}{\frac{1}{q} + q}$ $\Rightarrow s i n ϕ = \frac{2 b c}{a^{2} - b^{2} - c^{2}}$ $★$
	Commented by gsk2684 last updated on 24/Jun/21

	$thank you very much$
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