Question

In a triangle ABC , if sin A .sin(B-C)=sinC sin(A-B), then

• A. tan A, tan B, tan C are in A.P
• B. cot A, cot B, cot C, are in A.P
• C. cos 2A, cos2B, cos2C are in A.P
• D. sin 2A,sin2B,sin2C are in A.p

Answer 1

The correct option is C cos 2A, cos2B, cos2C are in A.P

In $△ABC,A+B+C=\pi$

Given $\sin A.\sin (B-C)=\sin C\sin (A-B)$

$⟹\sin (\pi -(B+C\left)\right)\sin (B-C)=\sin (\pi -(A+B\left)\right)\sin (A-B)$

$⟹\sin (B+C)\sin (B-C)=\sin (A+B)\sin (A-B)$

$⟹{\sin }^{2}B-{\sin }^{2}C={\sin }^{2}A-{\sin }^{2}B$

$⟹2{\sin }^{2}B-2{\sin }^{2}C=2{\sin }^{2}A-2{\sin }^{2}B$

$⟹(1-\cos 2B)-(1-\cos 2C)=(1-\cos 2A)-(1-\cos 2B)$

$⟹\cos 2C-\cos 2B=\cos 2B-\cos 2A$

$⟹2\cos 2B=\cos 2A+\cos 2C$

So $\cos 2A,\cos 2B,\cos 2C$ are in $A.P$