Question

Answer it.....
Please
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no. 58

Answer 1

Thanks for your patience.

Answer 2

Hope u like my process
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Formula to be used:-
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=> sec² x - tan²x = 1
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Given,
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$= > x = a \sec( \theta ) + b \tan( \theta )$
$= > y = a \tan( \theta ) + b \sec( \theta )$
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Now,

$= > {x}^{2} - {y}^{2} \\ \\ = {(a \sec( \theta ) + b \tan( \theta ) )}^{2} - {(a \tan( \theta ) + b \sec( \theta ) )}^{2} \\ \\ = {a}^{2} \sec ^{2} ( \theta ) + {b}^{2} \tan ^{2} ( \theta ) + 2ab \sec( \theta ) \tan( \theta ) - {a}^{2} \tan ^{2} ( \theta ) - {b}^{2} \sec ^{2} ( \theta ) - 2ab \sec( \theta ) \tan( \theta ) \\ \\ = {a}^{2} ( \sec ^{2} ( \theta ) - \tan ^{2} ( \theta ) ) - {b}^{2} ( \sec ^{2} ( \theta ) - \tan ^{2} ( \theta ) ) \\ \\ = {a}^{2} - {b}^{2} ....... < proved >$
Hope this is ur required answer

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