Hard MCQ +4 / -1 PYQ · JEE Mains 2025

The value of $ \cot^{-1} \left( \frac{\sqrt{1 + \tan^2(2)} - 1}{\tan(2)} \right) - \cot^{-1} \left( \frac{\sqrt{1 + \tan^2\left(\frac{1}{2}\right)} + 1}{\tan\left(\frac{1}{2}\right)} \right) $ is equal to

  1. A <p>$ \pi - \frac{3}{2} $</p>
  2. B <p>$ \pi + \frac{5}{2} $</p>
  3. C <p>$ \pi - \frac{5}{4} $</p> Correct answer
  4. D <p>$ \pi + \frac{3}{2} $</p>

Solution

<p>$\cot ^{-1}\left(\frac{|\sec 2|-1}{\tan 2}\right)-\cot ^{-1}\left(\frac{\left|\sec \frac{1}{2}\right|+1}{\tan \frac{1}{2}}\right)$</p> <p>$$\begin{aligned} & =\cot ^{-1}\left(\frac{-1-\cos 2}{\sin 2}\right)-\cot ^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right) \\ & =\pi-\cot ^{-1}(\cot 1)-\cot ^{-1}\left(\cot \frac{1}{4}\right) \\ & =\pi-1-\frac{1}{4}=\pi-\frac{5}{4} \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range

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