Let L is distance between two parallel normal | vedclass

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Question

$\operatorname{axsec} 	heta-\operatorname{bycosec} 	heta=a^{2}-b^{2}$

$\operatorname{axsec} 	heta-$ $bycosec$ $	heta=-\left(a^{2}-b^{2}ight)$

Vedclass · Accepted Answer

$2(a -b)$

Vedclass · Answer

$\operatorname{axsec} 	heta-\operatorname{bycosec} 	heta=a^{2}-b^{2}$

$\operatorname{axsec} 	heta-$ $bycosec$ $	heta=-\left(a^{2}-b^{2}ight)$

$L = \frac{{2\left( {{a^2} - {b^2}} ight)}}{{\sqrt {{a^2}{{\sec }^2}	heta  + {b^2}\cos e{c^2}	heta } }}$

$\because $ $\quad {a^2}{\sec ^2}	heta  + {b^2}\cos e{c^2}	heta  \ge {(a + b)^2}$

$\mathrm{L} \leq 2(\mathrm{a}-\mathrm{b})$

Let $L$ is distance between two parallel normals of , $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,\,a > b$ then maximum value of $L$ is

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