Question

`1/(c o s e ctheta-cottheta)-1/(sintheta)=1/(sintheta)-1/(c o s e ctheta+cottheta)`

Answer 1

We have
`LHS=(1)/(("cosec" theta - cot theta))-(1)/(sin theta) `
`=(1)/(("cosec" theta - cot theta)) xx (("cosec" theta + cot theta))/(("cosec" theta + cot theta)) -(1)/(sin theta) `
` =(("cosec" theta + cot theta))/(("cosec"^(2)theta - cot^(2)theta)) - "cosec" theta " "[because (1)/(sin theta)="cosec" theta] `
` = ("cosec" theta + cot theta)- "cosec" theta " "[because "cosec"^(2)theta - cot^(2)theta =1 ] `
` = cot theta. `
`RHS=(1)/(sin theta)-(1)/(("cosec" theta + cot theta))`
`="cosec" theta - (1)/(("cosec" theta + cot theta))xx(("cosec" theta - cot theta))/(("cosec" theta - cot theta))`
`= "cosec" theta - (("cosec" theta - cot theta))/(("cosec"^(2)theta - cot^(2)theta)) `
`="cosec" theta-("cosec" theta - cot theta) " " [ because "cosec"^(2)theta-cot^(2)theta=1] `
`= cot theta.`
` therefore LHS = RHS. `