prove without a calculator that if $\tan (p) = 1$ and $\tan(q) = 2$ then
$${\sin (p+q)\over \cos (p)\cos (q)} = 3$$
My attempt for the numerator: $\sin (p)\cos (q) + \cos (p)\sin (q)$ After this im stuck
$\endgroup$ 21 Answer
$\begingroup$$$\frac{\sin(p+q)}{\cos(p)\cos(q)}=\frac{\sin(p)\cos(q)+\sin(q)\cos(p)}{\cos(p)\cos(q)}=\tan(p)+\tan(q) \tag 1$$
Since $\tan(p)=1$ and $\tan(q)=2$, we find the result in $(1)$ to be $3$, as was to be shown.
$\endgroup$ 6
