In a right angled triangle the hypotenuse is | vedclass

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Question

$p^2sec^2	heta + p^2cosec^2	heta ={\left( {2\sqrt 2 } ight)^2} p^2$

$\Rightarrow$ $\frac{1}{{{{\sin }^2}	heta \,{{\cos }^2}	heta }}\,\,

Vedclass · Accepted Answer

$\frac{\pi }{8}$ & $\frac{3 \pi }{8}$

Vedclass · Answer

$p^2sec^2	heta + p^2cosec^2	heta ={\left( {2\sqrt 2 } ight)^2} p^2$

$\Rightarrow$ $\frac{1}{{{{\sin }^2}	heta \,{{\cos }^2}	heta }}\,\, = \,8$

$sin^22	heta = 1/2 = {\left( {\frac{1}{{\sqrt 2 }}} ight)^2}$

$2	heta = n\pi + \pi /4$

$	heta = n\pi /2 + \pi /8$

for $n = 0$      $\Rightarrow$      $	heta = \pi /8$

for $n=1$      $\Rightarrow$      $	heta = 3\pi /8$

In a right angled triangle the hypotenuse is $2 \sqrt 2$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

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