$\frac{cotx}{1+cosecx}+\frac{1+cosecx}{cotx}$ is equal to :
Answer & explanation
Correct answer: option 1
\(\frac{cotx}{1+cosecx}\) + \(\frac{1+cosecx }{cotx}\)
= \(\frac{cosx}{sinx+1}\) + \(\frac{sinx+1 }{cosx}\)
= \(\frac{cos²x + (1+sinx)²}{cosx(sinx+1)}\)
= \(\frac{cos²x + 1 +sin²x +2sinx}{cosx(sinx+1)}\)
{ sin²x + cos²x = 1 }
= \(\frac{2+2sinx}{cosx(sinx+1)}\)
= 2secx