If (A-B) A + ^2C^2A = 1, where A,B,C (0,2), then:

Question

If $\dfrac{	an(A-B)}{	an A} + \dfrac{\sin^2C}{\sin^2A} = 1$, where $A,B,C \in \left(0,\dfrac\pi2ight)$, then:

Accepted Answer

From the given relation: $\dfrac{\sin^2C}{\sin^2A} = 1 - \dfrac{\tan(A-B)}{\tan A} = \dfrac{\tan A - \tan(A-B)}{\tan A}$.

Using $\tan X - \tan Y = \dfrac{\sin(X-Y)}{\cos X\cos Y}$ with $X=A, Y=A-B$: $\tan A - \tan(A-B) = \dfrac{\sin B}{\cos A\cos(A-B)}$.

$$\frac{\sin^2C}{\sin^2A} = \frac{\sin B}{\cos A\cos(A-B)\tan A} = \frac{\sin B}{\sin A\cos(A-B)}$$

$$\sin^2C = \frac{\sin A\sin B}{\cos(A-B)} = \frac{\sin A\sin B}{\cos A\cos B + \sin A\sin B} = \frac{\tan A\tan B}{1+\tan A\tan B}$$

But also $\sin^2C = \dfrac{\tan^2C}{1+\tan^2C}$, so equating: $\dfrac{\tan^2C}{1+\tan^2C} = \dfrac{\tan A\tan B}{1+\tan A\tan B}$, which simplifies to:

$$\tan^2C = \tan A\tan B$$

Hence $\tan A, \tan C, \tan B$ are in G.P.

Solutions · 1