Question

f x=2 sec 0 and y = 3 cosec 0 then, find
the relation between x and ​y​

Answer 1

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

$sin ^{2} \alpha + cos ^{2} \alpha = 1$

by here we get

$\sin( \alpha ) = \sqrt{1 - \cos^{2} \alpha }$

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