If xsec@=1-ytan@ and x^2 sec^2@=5+y^2tan^2@ then x^2(y^2+4)/y^2=

Answer 1

Let @ = T

xsecT = 1 - ytanT

So, x^2(secT)^2 = 1 - 2ytanT + y^2(tanT)^2 = 5 + y^2(tanT)^2

2ytanT = -4

ytanT = -2

y = -2cotT

So, (y^2+4) / y^2 = [(4cotT)*2 + 4] / [4(cotT)^2]

= [(cotT)^2 + 1] / (cotT)^2 = (cscT)^2 / (cotT)^2 = (secT)^2

xsecT = 1 - ytanT = 1 - (-2cotT)tanT = 3

So, x = 3 / secT = 3cosT

Therefore, x^2(y^2+4) / y^2 = 9(cosT)^2(secT)^2 = 9