If S and S- be the foci of the ellipse x2-a2 - y2-b2 - 1 and e be its eccentricity- then tanPSS-4tanPS-S4 - 1 - e1 - e where P is any point on the ellipse-

The correct option is B FalseWe know that, PScosα +P S ^ ' cosβ =2ae #160; #160; #160; #160; (1)Similarly, by sine rule we can write, PS / ...

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Byju's Answer

Standard XII

Mathematics

Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

If S and ...

Question

If $S$ and $S^{'}$ be the foci of the ellipse $(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1$ and $e$ be its eccentricity, then $tan \frac{P S S^{'}}{4} tan \frac{P S^{'} S}{4} = \frac{1 - e}{1 + e}$ where $P$ is any point on the ellipse.

True

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False

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Solution

The correct option is B False
We know that,
$P S c o s α + P S^{^{'}} c o s β = 2 a e$ (1)

Similarly, by sine rule we can write,
$\frac{P S}{s i n β} = \frac{P S^{^{'}}}{s i n α}$

$∴ P S^{^{'}} = \frac{P S \times s i n α}{s i n β}$ (2)

Putting this value in equation (1), we get,
$P S c o s α + \frac{P S s i n α}{s i n β} \times c o s β = 2 a e$

$∴ P S (c o s α + \frac{s i n α}{s i n β} \times c o s β) = 2 a e$

$∴ P S (\frac{s i n β c o s α + s i n α c o s β}{s i n β}) = 2 a e$

$∴ P S (s i n β c o s α + s i n α c o s β) = 2 a e s i n β$

$∴ P S \times s i n (α + β) = 2 a e s i n β$

$∴ P S = \frac{2 a e s i n β}{s i n (α + β)}$

Put this value in equation (2), we get,
$P S^{^{'}} = \frac{2 a e s i n β}{s i n (α + β)} \times \frac{s i n α}{s i n β}$

$∴ P S^{^{'}} = \frac{2 a e s i n α}{s i n (α + β)}$

$P S + P S^{^{'}} = 2 a$

$∴ \frac{2 a e s i n β}{s i n (α + β)} + \frac{2 a e s i n α}{s i n (α + β)} = 2 a$

$∴ \frac{2 a e}{s i n (α + β)} (s i n α + s i n β) = 2 a$

$∴ \frac{e}{s i n (α + β)} (s i n α + s i n β) = 1$

$∴ e (s i n α + s i n β) = s i n (α + β)$

$∴ e (2 s i n (\frac{α + β}{2}) c o s (\frac{α - β}{2})) = 2 s i n (\frac{α + β}{2}) c o s (\frac{α + β}{2})$

$∴ e c o s (\frac{α - β}{2}) = c o s (\frac{α + β}{2})$

$∴ e = \frac{c o s (\frac{α + β}{2})}{c o s (\frac{α - β}{2})}$

$∴ \frac{1}{e} = \frac{c o s (\frac{α - β}{2})}{c o s (\frac{α + β}{2})}$

$∴ \frac{1 - e}{1 + e} = \frac{c o s (\frac{α - β}{2}) - c o s (\frac{α + β}{2})}{c o s (\frac{α - β}{2}) + c o s (\frac{α + β}{2})}$

$∴ \frac{1 - e}{1 + e} = \frac{2 s i n (\frac{α}{2}) s i n (\frac{β}{2})}{2 c o s (\frac{α}{2}) c o s (\frac{β}{2})}$

$∴ \frac{1 - e}{1 + e} = t a n (\frac{α}{2}) t a n (\frac{β}{2})$

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If S and S′ be the foci of the ellipse (x2/a2)+(y2/b2)=1 and e be its eccentricity, then tanPSS′4tanPS′S4=1−e1+e where P is any point on the ellipse.

If $S$ and $S^{'}$ be the foci of the ellipse $(x^{2} / a^{2}) + (y^{2} / b^{2}) = 1$ and $e$ be its eccentricity, then $tan \frac{P S S^{'}}{4} tan \frac{P S^{'} S}{4} = \frac{1 - e}{1 + e}$ where $P$ is any point on the ellipse.