fast answer plzz thanks
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Let x/a = sec∅cos∅
y/b = sec∅sin∅
z/c = tan∅
Then in
=x²/a² + y²/b²
=sec²∅cos²∅ + sec²∅sin²∅
=1/cos²∅ × cos²∅ + 1/cos²∅ × sin²∅ (as sec²∅=1/cos²)
=1 + tan²∅
So the ans would be 1 + tan²∅ = (1 + z²/c²)
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y/b = sec∅sin∅
z/c = tan∅
Then in
=x²/a² + y²/b²
=sec²∅cos²∅ + sec²∅sin²∅
=1/cos²∅ × cos²∅ + 1/cos²∅ × sin²∅ (as sec²∅=1/cos²)
=1 + tan²∅
So the ans would be 1 + tan²∅ = (1 + z²/c²)
:) HOPE THIS HELPS !!!!
LETS SUPPORT SYRIANS !!!!
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