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
$\frac{{\cos }^{2}33^{\circ }-\cos 57^{\circ }}{\cos 69^{\circ }-\cos 21^{\circ }}=\frac{\sin \left(57^{\circ }+33^{\circ }\right)\cdot \sin \left(57^{\circ }-33^{\circ }\right)}{-2\sin \frac{69+21}{2}\cdot \sin \frac{96-21}{2}}$
$\frac{\sin 90^{\circ }\times \sin 24^{\circ }}{-2\sin 45^{\circ }\sin 24^{\circ }}=\frac{1}{-2\times \frac{1}{\sqrt{2}}}=\frac{-1}{\sqrt{2}}$