First slide

Multiple and sub- multiple Angles

Question

If $(x - a) \cos θ + y \sin θ = (x - a) \cos ϕ + y \sin ϕ = a$ and $\tan (θ / 2) - \tan (ϕ / 2) = 2 b$ , then

Moderate

A

$y^{2} = 2 ax - (1 - b^{2}) x^{2}$
B

$\tan \frac{θ}{2} = \frac{1}{x} (y + bx)$
C

$y^{2} = 2 bx - (1 - a^{2}) x^{2}$
D

$\tan \frac{ϕ}{2} = \frac{1}{x} (y - bx)$

Solution

Let $\tan (θ / 2) = α and \tan (ϕ / 2) = β$ , so that $α - β = 2 b$ .
Also, $\cos θ = \frac{1 - \tan^{2} (θ / 2)}{1 + \tan^{2} (θ / 2)} = \frac{1 - α^{2}}{1 + α^{2}}$
and $\sin θ = \frac{2 \tan (θ / 2)}{1 + \tan^{2} (θ / 2)} = \frac{2 α}{1 + α^{2}}$
Similarly, $\cos ϕ = \frac{1 - β^{2}}{1 + β^{2}} and \sin ϕ = \frac{2 β}{1 + β^{2}}$
Therefore, we have from the given relations
$\begin{array}{l} (x - a) \frac{1 - α^{2}}{1 + α^{2}} + y (\frac{2 α}{1 + α^{2}}) = a \\ \Rightarrow {xα}^{2} - 2 yα + 2 a - x = 0 \end{array}$
Similarly ${xβ}^{2} - 2 yβ + 2 a - x = 0$
We see that $α and β$ are roots of the equation
${xz}^{2} - 2 yz + 2 a - x = 0$ ,
So that $α + β = 2 y / x and αβ = (2 a - x) / x$ .
Now, from $(α + β)^{2} = (α - β)^{2} + 4 αβ$ , we get
$\begin{array}{l} \Rightarrow {(\frac{2 y}{x})}^{2} = (2 b)^{2} + \frac{4 (2 a - x)}{x} \\ \Rightarrow y^{2} = 2 ax - (1 - b^{2}) x^{2} \end{array}$
Also, from $α + β = 2 y / x and α - β = 2 b$ , we get
$α = y / x + b and β = y / x - b$
$\Rightarrow \tan \frac{θ}{2} = \frac{1}{x} (y + bx) and \tan \frac{ϕ}{2} = \frac{1}{x} (y - bx)$

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