f x=2 sec 0 and y = 3 cosec 0 then, find
the relation between x and y
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Answer:
(xy)^2=9x^2+2y^2
Step-by-step explanation:
x=2sec A.
y=3cosecA
now start with any of one equation
I am starting with 1st eqn
x/2=secA
x/2=1/cosA
cosA = 2/x
we know the standard trigonometric identity
by here we get
here put the value of cos A
we get sinA as {√(x^2-2)}/x
now reciprocal of this will be equal to y/3 by 2nd equation
so we get
{√(x^2-2)}=3x/y
on squaring both sides
x^2-2 = 9x^2/y^2
on simplifying it we get
(xy)^2=9x^2+2y^2
if we replace A with x^2 and B with y^2
then we get
9A+2B=AB
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