Question

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

Answer 1

Answer:

Proof

Step-by-step 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

$(k+1)\frac{\cos(a)}{\sin(a)}= (k-1)\frac{\cos(b)}{\sin(b)}$ and finally

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