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]

Question

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]

Smart Achievers · Accepted Answer

(2)

We have,

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

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

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

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

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

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

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 $π + \frac{5}{2}$

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Solution

Q. The value of cot–1(1+tan2(2)–1tan(2))–cot–1(1+tan2(12)+1tan(12)) is equal to [2025]

1 π+32

2 π–54

3 π–32