Question

If sin (a+b)=k sin (a-b) prove that (k-1)cotb =( k+1) cota

Answer 1

Proof

Explanation:

Recall the identity
`\sin(a \pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b)`.

Consider
`\sin(a + b) = k\sin(a - b)`. We firstly apply the above identity to reach

`\sin(a)\cos(b) + \cos(a)\sin(b) = k(\sin(a)\cos(b) - \cos(a)\sin(b))`.

By expanding the bracket on the right we obtain

`\sin(a)\cos(b) + \cos(a)\sin(b) = k\sin(a)\cos(b) - k\cos(a)\sin(b)` and so

`\cos(a)\sin(b) + k\cos(a)\sin(b) = k\sin(a)\cos(b) - \sin(a)\cos(b)` and so

`(1+k)\cos(a)\sin(b) = (k-1)\sin(a)\cos(b)` and so

`(k+1)(\cos(a))/(\sin(a))= (k-1)(\cos(b))/(\sin(b))` and finally

`(k+1)\cot(a)= (k-1)\cot(b)`.