Prove the identity: (cot(^2)x/cscx-1) = (1+sinx/sinx)

Question

Show all the steps please.

Answer 1

If you look at the right hand side there are only sin terms, so you probably want to start by converting everything on the left hand side into sines and cosines

 

cot²x / (cscx - 1) = (cos²x/sin²x) / ( (1/sinx) - 1 )

 

The (1/sinx) - 1 in the denominator is annoying so rewrite it as (1 - sinx) / sinx 

 

(cos²x/sin²x) / ( (1-sinx)/sinx )

 

Perform the division by inverting the fraction in the denominator and multiplying

 

(cos²x/sin²x) * (sinx/(1-sinx)

 

When you have a term like 1-sinx it is sometimes useful to multiply by the conjugate.  So multiply the 2nd term by (1+sinx)/(1+sinx)

 

(cos²x/sin²x) * [ (sinx + sin²x) / (1-sin²x) ]

 

1-sin²x = cos²x so we are left with 

 

(sinx + sin²x) / sin²x = (1+sinx)/sinx  which is the right hand side