Question

What is $\frac{1-\tan 2^o\cot {62}^o}{\tan {152}^o-\cot {88}^o}$ equal to?

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

Answer 1

Answer: (B)

$-\sqrt{3}$

$\frac{1-\tan 2^0\cot {62}^0}{\tan {152}^0-\cot {88}^0}$

$=\frac{1-\tan 2^0\tan ({90}^0-{62}^0)}{\tan ({180}^0-{28}^0)-\tan ({90}^0-{88}^0)}$

$=\frac{1-\tan 2^0\tan {28}^0}{-\tan {28}^0-\tan 2^0}$

$=-\frac{1}{\left(\frac{\tan 2^0+\tan {28}^0}{1-\tan 2^0\tan {28}^0}\right)}$

$=-\frac{1}{\tan (2^0+{28}^0)}$ ...($tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$)

$=-\frac{1}{\tan {30}^0}$

$=-\sqrt{3}$

The answer is option (B)