Question

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

Answer 1

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

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric identity that

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

GIVEN

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

TO PROVE

$x² - y² = a²- b²$

PROOF

$x² - y²$

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

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

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

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

$= a² \times 1- b² \times 1$

$= a²- b²$

Hence proved

Answer 2

Answer:

Given

x = asecθ

Therefore x² = a²sec²θ

Similarly y = btanθ

Hence y² = b²tan²θ

Considering LHS,

x²/a² - y²/b²

= a²sec²θ / a² - b²tan²θ / b²

= sec²θ - tan²θ

= 1 + tan²θ - tan²θ

( Using property 1 + tan²θ = sec²θ )

= 1

= RHS

Therefore LHS = RHS

Hence proved