# Simplify, `cot10A.cot6A + cot10A.cot4A - cot6A.cot4A + 1 = 0`

### 1 Answer | Add Yours

`LHS = `

`= cot10A.cot6A+cot10A.cot4A-cot6A.cot4A+1`

`= cot10A(cot6A+cot4A)-cot6A.cot4A+1`

`= cot10A((cos6A)/(sin6A)+(cos4A)/(sin4A))-cot6A.cot4A+1`

`=cot10A((cos6A.sin4A+sin6A.cos4A)/(sin4A.sin6A))-(cos6A.cos4A)/(sin6A.sin4A)+1`

`= (cos10A)/(sin10A) xx (sin10A)/(sin4A.sin6A) - (cos6A.cos4A)/(sin6A.sin4A)+1`

`= (cos10A)/(sin4A.sin6A) - (cos6A.cos4A)/(sin6A.sin4A)+1`

`= (cos10A - cos6A.cos4A)/(sin6A.sin4A) + 1`

`= (cos(6A+4A) - cos6A.cos4A)/(sin6A.sin4A) + 1`

`= (cos6A.cos4A-sin6A.sin4A - cos6A.cos4A)/(sin6A.sin4A) + 1`

`= (-sin6A.sin4A)/(sin6A.sin4A) + 1`

`= -1+1`

`= 0`

Therefore,

`cot10A.cot6A+cot10A.cot4A-cot6A.cot4A+1 = 0`

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