Question

solve this

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prove that

$\tt sin \: \frac{\pi}{6} + cos^{2} \: \frac{\pi}{3} - tan^{2} \: \frac{\pi}{4} = \frac{1}{2}$






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Answer 1

Answer:

Solution

fXY(x,y)=∫∞−∞fXYZ(x,y,z)dz=∫1013(x+2y+3z)dz=13[(x+2y)z+32z2]10=13(x+2y+32),for0≤x,y≤1.

Thus,

fXY(x,y)=⎧⎩⎨⎪⎪13(x+2y+32)00≤x≤1,0≤y≤1otherwise

Answer 2

☞↬↬●$\large\sf \bf \underline\purple{ your\: \: answer \: !!}$

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