If `sin^(-1)((x^2-y^2)/(x^2+y^2))=loga ,t h e n(dy)/(dx)` is equal to `x/y` (b) `y/(x^2)` `(x^2-y^2)/(x^2+y^2)` (d) `y/x`

Options

A. `(x)/(y)`
B. `(y)/(x^(2))`
C. `(x^(2)-y^(2))/(x^(2)+y^(2))`
D. `(y)/(x)`

Explanation

`"We have "sin^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=log a`
`"or "(x^(2)-y^(2))/(x^(2)+y^(2))=sin (log a)`
`"or "(1-tan^(2)theta)/(1+tan^(2)theta)=sin(log a)" "("on putting "y= x tan theta)`
`"or "cos 2theta= sin (log a)`
`"or "2theta=cos^(-1)(sin (log a))`
`"or "theta=(1)/(2)cos^(-1)(sin (log a))`
`"or "tan^(-1)((y)/(x))=(1)/(2)cos^(-1)(sin (log a))`
`"or "(y)/(x)=tan ((1)/(2)cos^(-1)(sin (loga )))`
Differentiating w.r.t. x, we get
`(x(dy)/(dx)-y)/(x^(2))=0`
`"or "x(dy)/(dx)-y=0`
`"or "(dy)/(dx)=(y)/(x)`