Question

$\frac{\cos ^{2} 33^{\circ}-\cos ^{2} 57^{\circ}}{\sin 21^{\circ}-\cos 21^{\circ}}=$

• A. $\frac{1}{\sqrt{2}}$
• B. $-\frac{1}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{-\sqrt{3}}{2}$

Answer 1

Answer: (B)

Since $\cos ^{2} B-\cos ^{2} A=\sin (A+B) \cdot \sin (A-B)$
Hence, given expression can be written as
$\frac{\cos ^{2} 33^{\circ}-\cos 57^{\circ}}{\cos 69^{\circ}-\cos 21^{\circ}}=\frac{\sin \left(57^{\circ}+33^{\circ}\right) \cdot \sin \left(57^{\circ}-33^{\circ}\right)}{-2 \sin \frac{69+21}{2} \cdot \sin \frac{96-21}{2}}$
$\frac{\sin 90^{\circ} \times \sin 24^{\circ}}{-2 \sin 45^{\circ} \sin 24^{\circ}}=\frac{1}{-2 \times \frac{1}{\sqrt{2}}}=\frac{-1}{\sqrt{2}}$