If x = a sin theta , y= b tan theta , prove that a square / x square - b square / y square = 1
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Solution:-
x = a sin theta : y = b tan theta ....(Given)
L.H.S. = a²/x² - b²/y² = a²/(a sin theta)² - b²/(b tan theta)²
a² and b² are cancelled and then we get
1/sin² theta - 1/sin² theta/cos² theta ...[∴ tan theta = sin theta/cos theta]
1/sin² theta - cos² theta/sin² theta
= (1-cos² theta)/sin² theta ....[∴ sin² theta + cos² theta = 1]
sin² theta/sin² theta = 1
= R.H.S
⇒ a²/x² - b²/y²
Hence proved.
x = a sin theta : y = b tan theta ....(Given)
L.H.S. = a²/x² - b²/y² = a²/(a sin theta)² - b²/(b tan theta)²
a² and b² are cancelled and then we get
1/sin² theta - 1/sin² theta/cos² theta ...[∴ tan theta = sin theta/cos theta]
1/sin² theta - cos² theta/sin² theta
= (1-cos² theta)/sin² theta ....[∴ sin² theta + cos² theta = 1]
sin² theta/sin² theta = 1
= R.H.S
⇒ a²/x² - b²/y²
Hence proved.
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