Question

If
$P={\log }_2(\cos 1^{\circ }\cdot \cos 2^{\circ }\cdot \cos 4^{\circ }\cdot \cos 8^{\circ }\dots \cos {1024}^{\circ }\cdot \cos {2048}^{\circ })$
$Q={\log }_2\left((1+\tan {23}^{\circ })(1+\tan {22}^{\circ })\right)$
$R={\log }_2\left(\sin 1^{\circ }\right)$
and $S={\log }_2\left(\cos {46}^{\circ }\right),$
then the value of $|(P+Q+R)-S|$ equals

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

The correct option is C $11$

$P={\log }_2(\cos 1^{\circ }\cdot \cos 2^{\circ }\cdot \cos 4^{\circ }\cdot \cos 8^{\circ }\dots \cos {1024}^{\circ }\cdot \cos {2048}^{\circ })={\log }_2(\frac{1}{2^{12}}\cdot \frac{\sin {4096}^{\circ }}{\sin 1^{\circ }})={\log }_2(\frac{1}{2^{12}}\cdot \frac{\sin ({360}^{\circ }\times 11+{136}^{\circ })}{\sin 1^{\circ }})={\log }_2(\frac{1}{2^{12}}\cdot \frac{\sin {136}^{\circ }}{\sin 1^{\circ }})={\log }_2(\frac{1}{2^{12}}\cdot \frac{\sin {44}^{\circ }}{\sin 1^{\circ }})P={\log }_2\sin {44}^{\circ }-12-{\log }_2\sin 1^{\circ }$

$\because \tan ({22}^{\circ }+{23}^{\circ })=\frac{\tan {22}^{\circ }+\tan {23}^{\circ }}{1-\tan {22}^{\circ }\tan {23}^{\circ }}\Rightarrow (1+\tan {22}^{\circ })(1+\tan {23}^{\circ })=2\therefore Q={\log }_{2}2=1$

$R={\log }_2\sin 1^{\circ }$
and $S={\log }_2\cos {46}^{\circ }={\log }_2\sin {44}^{\circ }$

Now, $(P+Q+R)-S$
$={\log }_2\sin {44}^{\circ }-12-{\log }_2\sin 1^{\circ }+1+{\log }_2\sin 1^{\circ }-{\log }_2\sin {44}^{\circ }=-11$
$\therefore |(P+Q+R)-S|=11$