Question

a^2+b^2=2,c^2+d^2=1 then find the value of (ad-bc)^2+(ac+bd)^2 is

Answer 1

Answer :

$Given\:that,\\ \\ a^{2}+b^{2}=2,\:c^{2}+d^{2}=1\\ \\ Now,(ad-bc)^{2}+(ac+bd)^{2}\\ \\ =a^{2}d^{2}-2abcd+b^{2}c^{2}+a^{2}c^{2}+2abcd+b^{2}d^{2}\\ \\ =a^{2}d^{2}+b^{2}c^{2}+a^{2}c^{2}+b^{2}d^{2}\\ \\ =a^{2}d^{2}+a^{2}c^{2}+b^{2}c^{2}+b^{2}d^{2}\\ \\ =a^{2}(d^{2}+c^{2})+b^{2}(c^{2}+d^{2})\\ \\ =(c^{2}+d^{2})(a^{2}+b^{2})\\ \\=1\times2\\ \\=2$

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