If $\frac{sinx+cosx}{sinx-cosx}=\frac{6}{5}$ then the value of $\frac{tan^2x+1}{tan^2x-1}$ is :

$\frac{61}{60}$

Given :-

\(\frac{sinx + cosx   }{sinx - cosx}\) = \(\frac{6 }{5}\)

Applying componendo and dividendo

\(\frac{sinx + cosx +sinx - cos  }{sinx + cosx - sinx + cosx}\) = \(\frac{6+5 }{6-5}\)

\(\frac{2sinx  }{2 cosx}\) = \(\frac{11 }{1}\)

tanx = 11

on squaring both sides,

tan²x = 121

Now,

\(\frac{ tan²x + 1 }{tan²x - 1}\)

= \(\frac{121 + 1 }{121 - 1}\) 

= \(\frac{61 }{60}\)