The value of cot – 1 ( 1 + tan 2 ( 2 ) – 1 tan ( 2 ) ) – cot – 1 ( 1 + tan 2 ( 1 2 ) + 1 tan ( 1 2 ) ) is equal to [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The value of $\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$ is equal to [2025]

1 $π + \frac{3}{2}$

2 $π - \frac{5}{4}$

3 $π - \frac{3}{2}$

4 $π + \frac{5}{2}$

Ans.
(2)

We have,

$\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$

$= \cot^{- 1} (\frac{| \sec (2) | - 1}{\tan (2)}) - \cot^{- 1} (\frac{| \sec (\frac{1}{2}) | + 1}{\tan (\frac{1}{2})})$

$= \cot^{- 1} (\frac{- 1 - \cos 2}{\sin 2}) - \cot^{- 1} (\frac{1 + \cos (\frac{1}{2})}{\sin (\frac{1}{2})})$

$= \cot^{- 1} (\frac{- 2 \cos^{2} (1)}{2 \cos (1) \sin (1)}) - \cot^{- 1} (\frac{2 \cos^{2} (\frac{1}{4})}{2 \cos (\frac{1}{4}) \sin (\frac{1}{4})})$

$= \cot^{- 1} (\cot (- 1)) - \cot^{- 1} (\cot \frac{1}{4})$

$= π - 1 - \frac{1}{4} = π - \frac{5}{4}$ .

Want Us to Email you About Special Offers & Updates?