If `(sinx)(cosy)=1/2,t h e n(d^2y)/(dx^2)a t(pi/4,pi/4)` is `-4` (b) `-2` (c) `-6` (d) `0`

Options

A. `-4`
B. `-2`
C. `-6`
D. 0

Answer

A

Explanation

sin x cos `y =1/2`
Differentiating both the sides
`sin x(-siny) (dy)/(dx)+cos y cos x=0`
`(dy)/(dx) =(cos x cos y)/(sin x sin y)=cot x cot y`
When the point is `((pi)/(4),(pi)/(4))`
then `x=(pi)/(4) and y = (pi)/(4)`
`(dy)/(dx)=cot(pi)/(4).cot(pi)/(4)=1`
so `(d^(2)y)/(dx^(2)=cot(pi)/(4)(-cosec^(2)(pi)/(4))1+cot(pi)/(4)(-cosec^(2)(pi)/(4))`
`=1(-2)+1.(-2)=4`