Question

The value of ${\log }_{\cot {30}^{\circ }}\cot {60}^{\circ }$ is

• A. $0$
• B. $1$
• C. $-1$
• D. not defined

Answer 1

The correct option is C $-1$

${30}^{\circ }+{60}^{\circ }={90}^{\circ }$
Let $p=\cot {30}^{\circ }$ and
$q=\cot {60}^{\circ }$
$=\tan {30}^{\circ }$
$\Rightarrow pq=1$

$\therefore {\log }_{\cot {30}^{\circ }}\cot {60}^{\circ }={\log }_p\frac{1}{p}=-1$