If 1-Tan2°.Cot62°Tan152°-Cot88°=K3 then K=

1-tan2∘ cot62∘tan152∘-cot88∘ =1-*cot88∘ cot62∘-cot62∘-cot88∘=cot88∘ cot62∘-1cot 88∘+cot62∘=cot(88∘+62∘)=cot150∘=-3 ⇒k=-1

If 1-Tan2°.Cot62°Tan152°-Cot88°=K3 then K=

1-tan2∘ cot62∘tan152∘-cot88∘ =1-*cot88∘ cot62∘-cot62∘-cot88∘=cot88∘ cot62∘-1cot 88∘+cot62∘=cot(88∘+62∘)=cot150∘=-3 ⇒k=-1

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $\frac{1 - Tan 2 ° . Cot 62 °}{Tan 152 ° - Cot 88 °} = K \sqrt{3}$ then $K =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- 1$

$\frac{1}{2}$

$- \frac{1}{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{lcl} \frac{1 - \tan 2^{\circ} \cot 62^{\circ}}{\tan 152^{\circ} - \cot 88^{\circ}} \\ \begin{array}{cl} = & \frac{1 - * \cot 88^{\circ} \cot 62^{\circ}}{- \cot 62^{\circ} - \cot 88^{\circ}} \\ = & \frac{\cot 88^{\circ} \cot 62^{\circ} - 1}{\cot 88^{\circ} + \cot 62^{\circ}} \\ = & \cot (88^{\circ} + 62^{\circ}) \\ = & \cot 150^{\circ} = - \sqrt{3} \Rightarrow k = - 1 \end{array} \end{array}$