Maximum value of r where c2-r2-a2-sin2-b2-cos2- where c- a- b are constants is

The correct option is B c/a+b c^2r^2=a^2 #160; cosec^2θ+b^2 #160; sec^2θ 2c^2r^3drdθ= 2a^2 #160; cosec^2θ+2b^2 #160; sec^2θ #160; ...

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AM,GM,HM Inequality

Maximum value...

Question

Maximum value of $r$ where $\frac{c^{2}}{r^{2}} = \frac{a^{2}}{{sin}^{2} θ} + \frac{b^{2}}{{cos}^{2} θ}$ where $c, a, b$ are constants is

\frac{c}{a + b}

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\frac{c}{a - b}

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\frac{a^{2} + b^{2}}{c^{2}}

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\frac{c^{2}}{a^{2} + b^{2}}

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Solution

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\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

a, b, c

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{2} + b^{2} + c^{2}}{{(a + b + c)}^{2}} = \frac{a - b + c}{a + b + c}

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