Question

If $\frac{1-{tan2}^{\circ }{cot62}^{\circ }}{tan{152}^{\circ }-{cot88}^{\circ }}=k\sqrt{3},$the value of k is

• A. 1
• B. -1
• C. 1/2
• D. -1/2

Answer 1

Answer: (B)

-1

$\frac{1-\tan 2^{\circ }\cot {62}^{\circ }}{\tan {152}^{\circ }-\cot {88}^{\circ }}=k\sqrt{3}$

$\Rightarrow \frac{1-\tan 2^{\circ }\cot ({90}^{\circ }-{28}^{\circ })}{\tan ({180}^{\circ }-{28}^{\circ })-\cot ({90}^{\circ }-2^{\circ })}=k\sqrt{3}$

$\Rightarrow \frac{1-\tan 2^{\circ }\tan {28}^{\circ }}{-\tan {28}^{\circ }-\tan 2^{\circ }}=k\sqrt{3}$

$\Rightarrow -\frac{1-\tan 2^{\circ }\tan {28}^{\circ }}{-\tan {28}^{\circ }-\tan 2^{\circ }}=k\sqrt{3}$

$\Rightarrow \frac{-1}{\frac{\tan {28}^{\circ }+\tan 2^{\circ }}{1-\tan 2^{\circ }\tan {28}^{\circ }}}=k\sqrt{3}$

$\Rightarrow \frac{-1}{\tan ({28}^{\circ }+2^{\circ })}=k\sqrt{3}$

$\Rightarrow \frac{-1}{\tan {30}^{\circ }}=k\sqrt{3}$

$\Rightarrow -\cot {30}^{\circ }=k\sqrt{3}$

$\Rightarrow -\sqrt{3}=k\sqrt{3}$

$\therefore k=-1$

'