Question

if 3 angles A,B,C are in A.P then prove that
cotB=sinA-sinC/cosC-cosA?

Answer 1

Using the difference formula,

$\sin A -\sin C=2\sin(\frac{A-C}{2}) \cos (\frac{A+C}{2}) \\ \cos C -\cos A=2\sin(\frac{A-C}{2}) \sin (\frac{A+C}{2})$

Now,

$\frac{\sin A -\sin C}{\cos C -\cos A}=\frac{2\sin(\frac{A-C}{2}) \cos (\frac{A+C}{2}) }{2\sin(\frac{A-C}{2}) \sin (\frac{A+C}{2}) }\\ \frac{\sin A -\sin C}{\cos C -\cos A}=\cot (\frac{A+C}{2})$

Since A,B, C are in AP, $\frac{A+C}{2}=B$. Using this in the above equality,

$\frac{\sin A -\sin C}{\cos C -\cos A}=\cot (B)$

The proof is complete.

Answer 2

Answer:

Explanation: