If (a−b)sin⁡(θ+ϕ)=(a+b)sin⁡(θ−ϕ) and atan⁡θ2−btan⁡ϕ2=c then the value of sin Q is equal to

Since,(a−b)sin⁡(θ+ϕ)=(a+b)sin⁡(θ−ϕ)⇒a{sin⁡(θ+ϕ)−sin⁡(θ−ϕ)}=b{sin⁡(θ−ϕ)+sin⁡(θ+ϕ)}⇒2asin⁡ϕcos⁡θ=2bsin⁡θcos⁡ϕ⇒atan⁡ϕ=btan⁡θ⇒ 2atan⁡ϕ21−tan2⁡ϕ2=2btan⁡θ21−tan2⁡θ2⇒ 2atan⁡ϕ21−tan2⁡ϕ2=2btan⁡θ21−tan2⁡θ2 ……(i)Since, atan⁡θ2−btan⁡ϕ2=c (given)⇒ tan⁡θ2=btan⁡ϕ2+ca (ii)From Eqs. (i) and (ii), we get atan⁡ϕ21−tan2⁡ϕ2=btan⁡ϕ2+c1−btan⁡ϕ2+c2a2⇒ tan⁡ϕ2a2−b2−c2=bc1+tan2⁡ϕ2 Now, sin⁡ϕ=2tan⁡ϕ21+tan2⁡ϕ2=2bca2−b2−c2

If (a−b)sin⁡(θ+ϕ)=(a+b)sin⁡(θ−ϕ) and atan⁡θ2−btan⁡ϕ2=c then the value of sin Q is equal to

Since,(a−b)sin⁡(θ+ϕ)=(a+b)sin⁡(θ−ϕ)⇒a{sin⁡(θ+ϕ)−sin⁡(θ−ϕ)}=b{sin⁡(θ−ϕ)+sin⁡(θ+ϕ)}⇒2asin⁡ϕcos⁡θ=2bsin⁡θcos⁡ϕ⇒atan⁡ϕ=btan⁡θ⇒ 2atan⁡ϕ21−tan2⁡ϕ2=2btan⁡θ21−tan2⁡θ2⇒ 2atan⁡ϕ21−tan2⁡ϕ2=2btan⁡θ21−tan2⁡θ2 ……(i)Since, atan⁡θ2−btan⁡ϕ2=c (given)⇒ tan⁡θ2=btan⁡ϕ2+ca (ii)From Eqs. (i) and (ii), we get atan⁡ϕ21−tan2⁡ϕ2=btan⁡ϕ2+c1−btan⁡ϕ2+c2a2⇒ tan⁡ϕ2a2−b2−c2=bc1+tan2⁡ϕ2 Now, sin⁡ϕ=2tan⁡ϕ21+tan2⁡ϕ2=2bca2−b2−c2

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If $(a - b) \sin (θ + ϕ) = (a + b) \sin (θ - ϕ)$ and $atan \frac{θ}{2} - btan \frac{ϕ}{2} = c$ then the value of sin Q is equal to

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$\frac{2 ab}{a^{2} - b^{2} - c^{2}}$

$\frac{2 bc}{a^{2} - b^{2} - c^{2}}$

$\frac{2 bc}{a^{2} - b^{2} + c^{2}}$

$\frac{2 ab}{a^{2} - b^{2} + c^{2}}$

answer is B.

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Detailed Solution

Since, $(a - b) \sin (θ + ϕ) = (a + b) \sin (θ - ϕ)$

$\begin{array}{l} \Rightarrow a {\sin (θ + ϕ) - \sin (θ - ϕ)} = b {\sin (θ - ϕ) + \sin (θ + ϕ)} \\ \Rightarrow 2 asin ϕcos θ = 2 bsin θcos ϕ \Rightarrow atan ϕ = btan θ \\ \Rightarrow \frac{2 atan \frac{ϕ}{2}}{1 - \tan^{2} \frac{ϕ}{2}} = \frac{2 btan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}} \end{array}$

$\Rightarrow \frac{2 atan \frac{ϕ}{2}}{1 - \tan^{2} \frac{ϕ}{2}} = \frac{2 btan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}}$ ……(i)

Since, $atan \frac{θ}{2} - btan \frac{ϕ}{2} = c$ (given)

$\Rightarrow \tan \frac{θ}{2} = \frac{btan \frac{ϕ}{2} + c}{a}$ (ii)

From Eqs. (i) and (ii), we get

$\begin{array}{l} \frac{atan \frac{ϕ}{2}}{1 - \tan^{2} \frac{ϕ}{2}} = \frac{(btan \frac{ϕ}{2} + c)}{1 - \frac{{(btan \frac{ϕ}{2} + c)}^{2}}{a^{2}}} \\ \Rightarrow \tan \frac{ϕ}{2} (a^{2} - b^{2} - c^{2}) = bc (1 + \tan^{2} \frac{ϕ}{2}) \\ Now, \sin ϕ = \frac{2 \tan \frac{ϕ}{2}}{1 + \tan^{2} \frac{ϕ}{2}} = \frac{2 bc}{a^{2} - b^{2} - c^{2}} \end{array}$