UPSC NDA I 2025 Mathematics Paper

Given,
‌x= secθ−cos‌θ
‌y= sec4θ−cos4θ
We know from before:
(‌

dy

dx

)2=‌

16(y2+4)

x2+4

Differentiate both sides w.r.t. x :
‌

d

dx

[(x2+4)(‌

dy

dx

)2]=‌

d

dx

[16(y2+4)]
This yields the differential relation:
(x2+4)‌

d2y

dx2

+x‌

dy

dx

−16y=0
The required expression is
‌

x2+4

y2+4

‌

dy

dx

[(x2+4)‌

d2y

dx2

−16y]
From the differential relation,
(x2+4)‌

d2y

dx2

−16y=−x‌

dy

dx

So the expression becomes
‌

x2+4

y2+4

‌

dy

dx

(−x‌

dy

dx

)=−x‌

x2+4

y2+4

(‌

dy

dx

)2
Using (‌

dy

dx

)2=‌

16(y2+4)

x2+4

, all factors cancel except -16 x .
∴ The value of the given expression is -16 x .