If `sin^(-1)((x^2-y^2)/(x^2+y^2))=loga` then `(dy)/(dx)` is equal to (a)`(x^2-y^2)/(x^2+y^2)` (b) `y/x` (c) `x/y` (d) none of these

Explanation

We have given

`sin^(-1)((x^2-y^2)/(x^2+y^2))=loga`

`(x^2-y^2)/(x^2+y^2)=sin(loga)=lambda`, [`sin(loga)` is constant ]

`x^2−y^2=lambda(x^2+y^2)=lambdax^2+lambday^2`

`x^2(1−lambda)=y^2(1+lambda) ...(1)`

`⇒2x(1−lambda)=2y((dy)/(dx))(1+lambda)....(2)`

Dividing (2) by (1), we have

`(2/x)=(2/y)((dy)/(dx))⇒((dy)/(dx))=(y/x)`

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