# [Math] $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$, if $\sin(\alpha+\beta) = 1$ and $\sin(\alpha-\beta) = \frac 12$

trigonometry

If $\sin(\alpha+\beta) = 1$, $\sin(\alpha-\beta) = \frac 12$ , where $\alpha$, $\beta$ belongs to $[0,\frac \pi2],$ then what is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$?

What should be my approach to the question?

Just for the sake of a different approach -

We can make an observation first. If sin($\alpha$+$\beta$) = 1, then cos($\alpha$+$\beta$)=0; no matter what values $\alpha$ and $\beta$ take. Now, let me replace "$\alpha$+$\beta$" by another variable say "$A$". So, we have cos$A$ = 0. Therefore -

cos$3A$ = 4cos$^3A$ - 3cos$A$ = 4(0) - 3(0) = 0.

Thus, cos$3A$ = 0 = cos($3\alpha$+$3\beta$) ............ (1)

Now, writing our expression -

tan($\alpha$+$2\beta$)tan($2\alpha$+$\beta$)

= $\frac{2sin(\alpha+2\beta)sin(2\alpha+\beta)}{2cos(\alpha+2\beta)cos(2\alpha+\beta)}$

=$\frac{cos(\alpha-\beta) - cos(3\alpha+3\beta)}{cos(3\alpha+3\beta) + cos(\alpha-\beta)}$

=$\frac{cos(\alpha-\beta) - 0}{0 + cos(\alpha-\beta)}$ (from equation 1)

= 1