Question

`((cos A + cos B)/(sin A - sin B))^m`+ `((sin A + sin B)/(cos A - cos B))^m=_`?

Answer 1

Please see below.

Explanation:

`((cos A + cos B)/(sin A - sin B))^m`+ `((sin A + sin B)/(cos A - cos B))^m`

= `((2cos((A+B)/2)cos((A-B)/2))/(2cos((A+B)/2)sin((A-B)/2)))^m+((2sin((A+B)/2)cos((A-B)/2))/(-2sin((A+B)/2)sin((A-B)/2)))^m`

= `(cot((A-B)/2))^m+(-cot((A-B)/2))^m`

= `(cot((A-B)/2))^m+(-1)^m(cot((A-B)/2))^m`

As when `m` is odd `(-1)^m=-1`, and we have `((cos A + cos B)/(sin A - sin B))^m`+ `((sin A + sin B)/(cos A - cos B))^m=0`

and when `m` is even `(-1)^m=1` and we have `((cos A + cos B)/(sin A - sin B))^m`+ `((sin A + sin B)/(cos A - cos B))^m=2cot^m((A-B)/2)`

Answer 2

See the Proof given in the Explanation.

Explanation:

We have, `(cosA+cosB)/(sinA-sinB)`

`={2cos((A+B)/2)cos((A-B)/2)}/{2cos((A+B)/2)sin((A-B)/2)},`

`:. (cosA+cosB)/(sinA-sinB) =cot((A-B)/2)=x, say, ......(1),` where,

`x=cot((A-B)/2).`

Likewise, `(sinA+sinB)/(cosA-cosB)=-cot((A-B)/2)=-x, ........(2).`

`:. {(cosA+cosB)/(sinA-sinB)}^m+{(sinA+sinB)/(cosA-cosB)}^m, m in NN,`

`=x^m+(-x)^m,`

`=x^m+(-1)^mx^m,`

`={(-1)^m+1}x^m,`

`={(-1)^m+1}cot^m((A-B)/2), m in NN.`

Case 1 : `m in NN, m" is odd."`

`:. (-1)^m+1=-1+1=0.`

`rArr {(cosA+cosB)/(sinA-sinB)}^m+{(sinA+sinB)/(cosA-cosB)}^m, m in NN, m" is odd",`

`={(-1)^m+1}cot^m((A-B)/2), m in NN, m" is odd,"`

`=0.`

Case 2 : `m in NN, m" is even."`

Here, `(-1)^m+1=1+1=2.`

`rArr {(cosA+cosB)/(sinA-sinB)}^m+{(sinA+sinB)/(cosA-cosB)}^m, m in NN, m" is even",`

`={(-1)^m+1}cot^m((A-B)/2), m in NN, m" is even,"`

`=2cot^m((A-B)/2).`

Hence, the Proof.

Enjoy Maths.!