$$\sin^2 21-\sin^2 9=(\sin 21-\sin 9)(\sin 21+\sin 9)$$
$$=$$
$$2 \cos\left(\frac{21+9}{2}\right)\sin\left(\frac{21-9}{2}\right)2\sin\left(\frac{21+9}{2}\right) \cos\left(\frac{21-9}{2}\right) $$
$$=$$
$$2.\cos15.\sin6.2\sin15. \cos6$$
$$=$$
$$\sin6.\cos6$$
$\tan6=a$ olduğuna göre gerisi kolay.