JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestionThe value of cot−1(1+tan2(2)−1tan(2))−cot−1(1+tan2(12)+1tan(12))\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)cot−1(tan(2)1+tan2(2)−1)−cot−1(tan(21)1+tan2(21)+1) is equal toOptionsAπ−32\pi - \frac{3}{2}π−23Bπ+52\pi + \frac{5}{2}π+25Cπ−54\pi - \frac{5}{4}π−45Dπ+32\pi + \frac{3}{2}π+23Check AnswerHide SolutionSolutioncot−1(∣sec2∣−1tan2)−cot−1(∣sec12∣+1tan12)\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)cot−1(tan2∣sec2∣−1)−cot−1(tan21∣sec21∣+1) =cot−1(−1−cos2sin2)−cot−1(1+cos12sin12)=π−cot−1(cot1)−cot−1(cot14)=π−1−14=π−54\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}=cot−1(sin2−1−cos2)−cot−1(sin211+cos21)=π−cot−1(cot1)−cot−1(cot41)=π−1−41=π−45