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

$\cot 16 ° \cot 44 ° + \cot 44 ° \cot 76 ° - \cot 16 ° =$

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

\cot 16 ° \cot 44 ° + \cot 44 ° \cot 76 ° - \cot 76 ° \cot 16 °

$= \tan 74 ° . \tan 46 ° + \tan 46 ° . \tan 14 ° - \tan 14 ° . \tan 74 °$

Using $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A . \tan B} \Rightarrow \tan A . \tan B = 1 - \frac{\tan A + \tan B}{\tan (A + B)}$

$\Rightarrow \tan 74 ° . \tan 46 ° = 1 - \frac{\tan 74 ° + \tan 46 °}{\tan (74 + 46)} = 1 - \frac{\tan 74 ° + \tan 46 °}{\tan 120 °}$

$= 1 + \frac{\tan 74 ° ° + \tan 46 °}{\sqrt{3}} \to (1)$

$\tan 46 ° . \tan 14 ° = 1 - \frac{\tan 46 ° + \tan 14 °}{\tan (46 ° + 14 °)} = \frac{1 - \tan 46 ° + \tan 14 °}{\tan 60 °}$

$= 1 - \frac{\tan 46 ° + \tan 14 °}{\sqrt{3}} \to (2)$

Using $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A . \tan B} \Rightarrow \tan A . \tan B = \frac{\tan A - \tan B}{\tan (A - B)} - 1$

$\tan 74 ° . \tan 14 ° = \frac{\tan 74 ° - \tan 14 °}{\tan (74 ° - 14 °)} - 1 = \frac{\tan 74 ° - \tan 14 °}{\tan 60 °} - 1$

$= \frac{\tan 74 ° - \tan 14 °}{\sqrt{3}} - 1 \to (3)$

From (1), (2) and (3)

$\tan 74 ° . \tan 46 ° + \tan 46 ° . \tan 14 ° - \tan 74 ° . \tan 14 °$

$= 1 + \frac{\tan 74 ° + \tan 46 °}{\sqrt{3}} + 1 - \frac{\tan 46 ° + \tan 14 °}{\sqrt{3}} - \frac{\tan 74 ° - \tan 14 °}{\sqrt{3}} + 1$

$= 3$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!