If 2ycosA=xsinA and 2xsecA-ycosecA=3, then find the value of x²+4y².

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Solution

2y cosA= xsinA
cosA=xsinA/2y

since, cosA=1/secA
therefore, secA= 1/cosA
2y/xsinA

now put the value of secA in second equation
2x(2y/xsinA) - ycosecA=3
4y/sinA - y/sinA=3 since cosecA=1/sinA
3y/sinA = 3
y= sinA.
from equation 1,
x= 2sinAcosA/sinA
x= 2cosA

x^2+4y^2=(2cosA)^2+4(sinA)^2
=4cos^2A+4sin^2A
=4(cos^2A+sin^2A)
=4.
x^2+4y^2=4.

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