Question

Evaluate $\cos {48}^0-\sin {42}^0$

Answer 1

$\cos {48}^0=\cos ({90}^0-{42}^0)$

$=\sin {42}^0$ $[\cos ({90}^0-\mathrm{\theta })=\mathrm{sin\theta }]$

⇒$\cos {48}^0-\sin {42}^0$ $=\sin {42}^0$$-\sin {42}^0$

$=0$

Thus, $\cos {48}^0-\sin {42}^0$$=0$ Answer