Question

The value of $\frac{{\cos }^2{33}^{\circ }-{\cos }^2{57}^{\circ }}{\sin {21}^{\circ }-\cos {21}^{\circ }}$ is

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

Answer 1

The correct option is A $\frac{-1}{\sqrt{2}}$

Using trignometric rule:
$\cos \theta =\sin (90-\theta )$
By using this, we get
$\frac{{\cos }^{2}33-{\cos }^{2}57}{\sin 21-\cos 21}=\frac{{\cos }^{2}33-{\sin }^2(90-57)}{\sin 21-\sin (90-21)}=\frac{{\cos }^{2}33-{\sin }^{2}33}{\sin 21-\sin 69}$
Now by using trignometric rule,
${\cos }^{2}A-{\sin }^{2}A=\cos 2A$ and $\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$
Thus $\frac{{\cos }^{2}33-{\sin }^{2}33}{\sin 21-\sin 69}=\frac{\cos 66}{2\cos 45\sin 24}$

$=\frac{\sin (90-66)}{2\frac{1}{\sqrt{2}}\sin 24}=\frac{\sin 24}{\sqrt{2}\sin 24}=\frac{1}{\sqrt{2}}$