If $\frac{sinx+cosx}{sinx-cosx}=\frac{6}{5}$ then the value of $\frac{tan^2x+1}{tan^2x-1}$ is :
Answer & explanation
Correct answer: option 2
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,
\(\frac{ tan²x + 1 }{tan²x - 1}\)
= \(\frac{121 + 1 }{121 - 1}\)
= \(\frac{61 }{60}\)