Question

If x is equal to 7 + 4 root 3 and xy is equal to 1 then 1 upon x square + 1upon y^2 square is equal to

Answer 1

if u found any difficulty please ask in comment box ..I would like to help you for this question and any other question

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: Given, \: }} \\ \\ \: \: \: \: \: \: \tt{ x = 7 + 4 \sqrt{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: \tt{ xy = 1} \\ \\ \underline{ \mathfrak{ \:To \: \: find :- \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \tt {\frac{1}{x {}^{2} } + \frac{1}{y {}^{2} } = \: ?}$

$\underline{ \mathfrak{ \: Now, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \tt{xy = 1 }\\ \\ \tt{\implies y = \frac{1}{x} } \\ \\ \tt{\implies y = \frac{1}{7 +4 \sqrt{3} }} \\ \\ \tt{ \implies y = \frac{1(7 - 4 \sqrt{3} )}{(7 + 4 \sqrt{3})(7 - 4 \sqrt{3}) }} \\ \\ \tt{\implies y = \frac{7 - 4 \sqrt{3} }{(7) {}^{2} - (4 \sqrt{3} ) {}^{2} }} \\ \\ \tt{\implies y = \frac{7 - 4 \sqrt{3} }{49 - 48} } \\ \\ \tt{\implies y = \frac{7 - 4 \sqrt{3} }{1}} \\ \\ \: \: \: \: \: \: \tt{ \therefore \: \: y= 7 - 4 \sqrt{3} }$

$\underline{ \mathfrak{ \: Again, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf{xy = 1} \\ \\ \sf{ \implies \: y = \frac{1}{x} } \\ \\ \sf{ \implies \: y {}^{2} = ( \frac{1}{x}) {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{x {}^{2} } = y {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{x {}^{2} } = (7 - 4 \sqrt{3} ) {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{x {}^{2} } = (7) {}^{2} - 2 \times 7 \times 4\sqrt{3} + (4 \sqrt{3} ) {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{x {}^{2} } =49 - 56 \sqrt{3} + 48 } \\ \\ \: \: \: \: \: \: \: \: \sf{ \therefore \: \: \underline{ \: \: \: \frac{1}{x {}^{2} } =97 - 56 \sqrt{3} \: \: \: }}$

$\underline{\mathfrak{ \: And, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf{xy = 1} \\ \\ \sf{ \implies \: x = \frac{1}{y} } \\ \\ \sf{ \implies \: x {}^{2} = ( \frac{1}{y}) {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{y {}^{2} } = x {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{y {}^{2} } = (7 + 4 \sqrt{3} ) {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{y {}^{2} } = (7) {}^{2} + 2 \times 7 \times 4\sqrt{3} + (4 \sqrt{3} ) {}^{2} } \\ \\ \sf{ \implies \: \frac{1}{y {}^{2} } =49 + 56 \sqrt{3} + 48 } \\ \\ \: \: \: \: \: \: \: \: \sf{ \therefore \: \: \underline{ \: \: \: \frac{1}{y {}^{2} } =97 + 56 \sqrt{3} \: \: \: \: } \: }$

$\bold{ \therefore \: \: \red{ \frac{1}{x {}^{2} } + \frac{1}{y {}^{2}}} = (97 - 56 \sqrt{3} ) + (97 + 56\sqrt{3} )} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 97 - \cancel{ 56 \sqrt{3}} + 97 + \cancel{56\sqrt{3}}} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 97 + 97} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \red{194}}$