It is known that Cartesian to Polar is x= r#cos theta# and y= r #sin theta#
Hence dx=# (dr)/(d theta) cos theta -r sin theta# and
dy= #(dr)/(d theta) sin theta +r cos theta#
Therefore slope #dy/dx =((dr)/(d theta) sin theta + r cos theta)/((dr)/(d theta) cos theta - r sin theta#
When #theta= pi/4# , slope would be =# ((dr)/(d theta) +r)/((dr)/(d theta) -r )# , because for # theta= pi/4#, #sin theta = cos theta #
Now given r= #tan^2 theta -cos theta#, #(dr)/(d theta ) =2 tan theta sec^2 theta + sin theta #. For #theta = pi/4#, #(dr)/(d theta)= 2*1*2 + 1/sqrt2# = #4+ 1/sqrt2# and r= 1- #1/sqrt2#
Hence slope= #5/(3+2/sqrt2)# = # 5/(3+sqrt2)#
