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

Question

If $\frac{	an (A-B)}{	an A}+\frac{{\sin }^{2}C}{{\sin }^{2}A}=1,A,B,C\in (0,\frac{\pi }{2})$, then

Accepted Answer

Given :

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

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

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

We know, $\tan X-\tan Y=\frac{\sin (X-Y)}{\cos X\cos Y}$

$\therefore $ $\tan A-\tan (A-B)=\frac{\sin (A-(A-B))}{\cos A\cos (A-B)}=\frac{\sin B}{\cos A\cos (A-B)}$

Substitute back into (1)

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

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

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

Divide by $\cos A\cos B$ :

${\sin }^{2}C=\frac{\tan A\tan B}{\frac{\cos A\cos B+\sin A\sin B}{\cos A\cos B}}$

$\Rightarrow$ ${\sin }^{2}C=\frac{\tan A\tan B}{1+\tan A\tan B}$

$\Rightarrow$ ${\sin }^{2}C=\frac{{\tan }^{2}C}{1+{\tan }^{2}C}$

$\Rightarrow$ $\frac{{\tan }^{2}C}{1+{\tan }^{2}C}=\frac{\tan A\tan B}{1+\tan A\tan B}$

$\Rightarrow$ ${\tan }^{2}C+{\tan }^{2}C\tan A\tan B=\tan A\tan B+{\tan }^{2}C\tan A\tan B$

$\Rightarrow$ ${\tan }^{2}C=\tan A\tan B$

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

Solutions · 1