# tan(A-B)/tan(A) + (sinC)^2/(sinA)^2=1

• December 27th 2009, 08:13 PM
sovan
tan(A-B)/tan(A) + (sinC)^2/(sinA)^2=1
If tan(A-B)/tan(A) + {sin(C)}^2 / {sin(A)}^2 = 1
Then
Show that
tan(A).tan(B)={tan(C)}^2

How can it be done?

Thanks
• December 27th 2009, 11:19 PM
Quote:

Originally Posted by sovan
tan(A).tan(B)={tan(C)}^2

tan(A) = {tan(x)}^2/tan(B)

Replace in the initial equation .
• December 28th 2009, 12:29 AM
Bacterius
Quote:

tan(A) = {tan(x)}^2/tan(B)

Replace in the initial equation .

Substituting into the original equation should work. If you could use LaTeX tags next time, it sort of helps :

Quote:

If $\frac{\tan{(A-B)}}{\tan{(A)}} + \frac{\sin{(C)}^2}{\sin{(A)}^2} = 1$
Then
Show that
$\tan{(A)} \cdot \tan{(B)} = \tan{(C)}^2$

How can it be done?

Thanks