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Geometry Level 2

( csc ⁡ A csc ⁡ B + cot ⁡ A cot ⁡ B ) 2 − ( csc ⁡ A cot ⁡ B + csc ⁡ B cot ⁡ A ) 2 = ? (\csc A \csc B+\cot A \cot B)^2 - (\csc A \cot B+\csc B \cot A)^2 = \ ?

This problem is part of this set .
2 -1 0 1

Vaibhav Kandwal by Vaibhav Kandwal , 21, India

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1 solution

Marvin Chong
Feb 15, 2020

If you open up all the brackets, the ' 2 x y 2xy ' of the perfect square identity will cancel out, thus, you are left with:

csc ⁡ 2 A csc ⁡ 2 B + cot ⁡ 2 A cot ⁡ 2 B − csc ⁡ 2 A cot ⁡ 2 B − csc ⁡ 2 B cot ⁡ 2 A \csc ^{ 2 }{ A } \csc ^{ 2 }{ B } +\cot ^{ 2 }{ A } \cot ^{ 2 }{ B } -\csc ^{ 2 }{ A } \cot ^{ 2 }{ B } -\csc ^{ 2 }{ B } \cot ^{ 2 }{ A }

If you factor them as so:

csc ⁡ 2 A ( csc ⁡ 2 B − cot ⁡ 2 B ) + cot ⁡ 2 A ( cot ⁡ 2 B − csc ⁡ 2 B ) \csc ^{ 2 }{ A } (\csc ^{ 2 }{ B } -\cot ^{ 2 }{ B } )+\cot ^{ 2 }{ A } (\cot ^{ 2 }{ B } -\csc ^{ 2 }{ B } ) ⇒ csc ⁡ 2 A ( csc ⁡ 2 B − cot ⁡ 2 B ) − cot ⁡ 2 A ( csc ⁡ 2 B − cot ⁡ 2 B ) \Rightarrow \csc ^{ 2 }{ A } (\csc ^{ 2 }{ B } -\cot ^{ 2 }{ B } )-\cot ^{ 2 }{ A } (\csc ^{ 2 }{ B } -\cot ^{ 2 }{ B } ) ⇒ ( csc ⁡ 2 A − cot ⁡ 2 A ) ( csc ⁡ 2 B − cot ⁡ 2 B ) \Rightarrow (\csc ^{ 2 }{ A } -\cot ^{ 2 }{ A } )(\csc ^{ 2 }{ B } -\cot ^{ 2 }{ B } )

Since csc ⁡ 2 x = 1 + cot ⁡ 2 x \csc ^{ 2 }{ x } =1+\cot ^{ 2 }{ x } , csc ⁡ 2 x − cot ⁡ 2 x = 1 \csc ^{ 2 }{ x } -\cot ^{ 2 }{ x } =1 , therefore, the answer is 1 \boxed{ 1 } .

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