# cotan(a+b)cotan(a-b) = ?As there is sin(a+b)sin(a-b) = sin²a - sin²b ...

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should use the formula `cot alpha = cos alpha/sin alpha` , such that:

`cot(a+b)*cot(a-b) = (cos(a+b)cos(a-b))/(sin(a+b)*sin(a-b))`

`(cos(a+b)cos(a-b)) = (1/2)(cos(a+b+a-b) + cos(a+b-a+b))`

`(sin(a+b)*sin(a-b)) =(1/2)(cos(a+b-a+b) - cos(a+b+a-b))`

You need to substitute `(1/2)(cos(a+b+a-b) + cos(a+b-a+b))`  for `(cos(a+b)cos(a-b))`  and `(1/2)(cos(a+b-a+b) - cos(a+b+a-b)) ` for `(sin(a+b)*sin(a-b))`  such that:

`cot(a+b)*cot(a-b) = ((1/2)(cos(a+b+a-b) + cos(a+b-a+b)))/((1/2)(cos(a+b-a+b) - cos(a+b+a-b)))`

`cot(a+b)*cot(a-b) = (cos(2a) + cos(2b))/(cos(2b) - cos(2a))`

You may use the formula for cosine of double angle such that:

`cot(a+b)*cot(a-b) = (cos^2 a - sin^2 a+ cos^2 b - sin^2 b)/(cos^2 b - sin^2 b- cos^2 a + sin^2 a)`

Hence, evaluating the product `cot(a+b)*cot(a-b)`  yields `cot(a+b)*cot(a-b)= (cos(2a) + cos(2b))/(cos(2b) - cos(2a)).`