The value of $\frac{\sin 30^{\circ} \sin 60^{\circ}}{\cos 60^{\circ} \cos 30^{\circ}}-\tan 45^{\circ}$ is:
Answer & explanation
Correct answer: option 2
\(\frac{sin30º. sin60º}{cos60º.cos30º}\) - tan45º
{ using sinA = cosB Iff A + B = 90º }
= \(\frac{sin30º. sin60º}{sin30º.sin60º}\) - tan45º
= \(\frac{1}{1}\) - 1
= 1