Question

If x = a sec A + b tan A and y = a tan A + b sec A, prove that x² - y² = a² - b².


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Answer 1

Solution :

Given :

• x = a•secA + b•tanA

• y = a•tanA + b•secA

To prove :

• x² - y² = a² - b²

Proof :

LHS

= x² - y²

= (x + y) × (x - y)

= [ (a•secA + b•tanA) + (a•tanA + b•secA) ]

× [ (a•secA + b•tanA) - (a•tanA + b•secA) ]

= [ a•secA + b•tanA + a•tanA + b•secA ]

× [ a•secA + b•tanA - a•tanA - b•secA ]

= [ a•(secA + tanA) + b•(secA + tanA) ]

× [ a•(secA - tanA) - b•(secA - tanA) ]

= (secA + tanA)(a + b) × (seA - tanA)(a - b)

= (a + b)(a - b) × (secA + tanA)(secA - tanA)

= (a² - b²) × (sec²A - tan²A)

= (a² - b²) × 1 { °•° sec²A - tan²A = 1 }

= a² - b²

= RHS

Hence proved .

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

${ \sec}^{2} A - { \tan}^{2} A = 1$

GIVEN

$x = a \sec \: A + b \tan \: A \: \: and \: \: y = a \tan \: A \: + b \sec \: A$

TO PROVE

${x}^{2} - {y}^{2} = {a}^{2} - {b}^{2}$

PROOF

${x}^{2} - {y}^{2}$

$= {( a \sec \: A + b \tan \: A)}^{2} \: - {(a \tan \: A \: + b \sec \: A )}^{2} \: \:$

$= ({a}^{2} { \sec}^{2} A + 2ab \: \sec A \tan A \: + {b}^{2} { \tan}^{2} A) - ({a}^{2} { \tan}^{2} A + 2ab \: \sec A \tan A \: + {b}^{2} { \sec}^{2} A)$

$= {a}^{2} { \sec}^{2} A + 2ab \: \sec A \tan A \: + {b}^{2} { \tan}^{2} A - {a}^{2} { \tan}^{2} A - 2ab \: \sec A \tan A \: - {b}^{2} { \sec}^{2} A$

$= {a}^{2} { \sec}^{2} A + {b}^{2} { \tan}^{2} A - {a}^{2} { \tan}^{2} A\: - {b}^{2} { \sec}^{2} A$

$= {a}^{2} { \sec}^{2} A - {a}^{2} { \tan}^{2} A\: - {b}^{2} { \sec}^{2} A \: + {b}^{2} { \tan}^{2} A$

$= {a}^{2} ( { \sec}^{2} A - { \tan}^{2} A\:) - {b}^{2} ( { \sec}^{2} A \: - { \tan}^{2} A)$

$= {a}^{2} \times 1 - {b}^{2} \times 1$

$= {a}^{2} - {b}^{2}$

Hence proved