The slope of the tangent at `(x , y)` to a curve passing through `(1,pi/4)` is given by `y/x-cos^2(y/x),` then the equation of the curve is (a) `( b ) (c) y=( d ) (e)tan^(( f ) (g)-1( h ))( i )(( j ) (k)log(( l ) (m) (n) e/( o ) x (p) (q) (r))( s ))( t )` (u) (v) `( w ) (x) y=x (y) (z)tan^(( a a ) (bb)-1( c c ))( d d )(( e e ) (ff)log(( g g ) (hh) (ii) x/( j j ) e (kk) (ll) (mm))( n n ))( o o )` (pp) (qq) `( r r ) (ss) y=x (tt) (uu)tan^(( v v ) (ww)-1( x x ))( y y )(( z z ) (aaa)log(( b b b ) (ccc) (ddd) e/( e e e ) x (fff) (ggg) (hhh))( i i i ))( j j j )` (kkk) (d) none of these

Options

A. `y=tan^(-1) [ log(e/x)]`
B. `y=x tan^(-1) [log (x/e)]`
C. `y=x tan^(-1) [log (e/x)]`
D. None of these

Answer

C