fast answer plzz thanks
Attachments:

Answers
Answered by
0
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)
:) HOPE THIS HELPS !!!!
LETS SUPPORT SYRIANS !!!!
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 !!!!
nitthesh7:
PLS SUPPORT SYRIANS LOOK AT THEIR DIFFICULTIES IN GOOGLE LETS SUPPORT SYRIANS LET THE INSPIRATION START FROM US
Similar questions