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Inverse Trigonometric Functions question

2025 · 8 Apr · Shift 2 · Q34
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  5. /2025 · 8 Apr · Shift 2 · Q34

Inverse Trigonometric Functions question

2025 · 8 Apr · Shift 2 · Q34

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The value of cot⁡−1(1+tan⁡2(2)−1tan⁡(2))−cot⁡−1(1+tan⁡2(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 to
  1. A
    π−32\pi - \frac{3}{2}π−23​
  2. B
    π+52\pi + \frac{5}{2}π+25​
  3. C
    π−54\pi - \frac{5}{4}π−45​
  4. D
    π+32\pi + \frac{3}{2}π+23​
View written solutionFree

Correct answer: C

  1. Simplify the first inverse cotangent

Let x=2.x=2.x=2. Then 1+tan⁡2x=∣sec⁡x∣.\sqrt{1+\tan^2 x}=|\sec x|.1+tan2x​=∣secx∣. Since 2∈(π2,π)2\in\left(\frac{\pi}{2},\pi\right)2∈(2π​,π), we have cos⁡2<0\cos 2<0cos2<0, so ∣sec⁡2∣=−sec⁡2.|\sec 2|=-\sec 2.∣sec2∣=−sec2. Thus \frac{\sqrt{1+\tan^2 2}-1}{\tan 2}= rac{-\sec 2-1}{\tan 2}. Now use tan⁡x2=sec⁡x−1tan⁡x=tan⁡xsec⁡x+1.\tan\frac{x}{2}=\frac{\sec x-1}{\tan x}=\frac{\tan x}{\sec x+1}.tan2x​=tanxsecx−1​=secx+1tanx​. Hence sec⁡x+1tan⁡x=cot⁡x2.\frac{\sec x+1}{\tan x}=\cot\frac{x}{2}.tanxsecx+1​=cot2x​. So −sec⁡2−1tan⁡2=−sec⁡2+1tan⁡2=−cot⁡1.\frac{-\sec 2-1}{\tan 2}=-\frac{\sec 2+1}{\tan 2}=-\cot 1.tan2−sec2−1​=−tan2sec2+1​=−cot1. Therefore the first term is cot⁡−1(−cot⁡1).\cot^{-1}(-\cot 1).cot−1(−cot1).

Taking the principal value range of cot⁡−1\cot^{-1}cot−1 as (0,π)(0,\pi)(0,π), and since 1∈(0,π)1\in(0,\pi)1∈(0,π), cot⁡−1(−cot⁡1)=π−1.\cot^{-1}(-\cot 1)=\pi-1.cot−1(−cot1)=π−1.

So, T1=π−1.T_1=\pi-1.T1​=π−1.

  1. Simplify the second inverse cotangent

Let y=12.y=\frac12.y=21​. Since 12∈(0,π2)\frac12\in\left(0,\frac{\pi}{2}\right)21​∈(0,2π​), we have cos⁡12>0\cos\frac12>0cos21​>0, so 1+tan⁡212=sec⁡12.\sqrt{1+\tan^2\frac12}=\sec\frac12.1+tan221​​=sec21​. Thus

=\frac{\sec\frac12+1}{\tan\frac12}.$$ Using the identity $$\frac{\sec y+1}{\tan y}=\cot\frac{y}{2},$$ we get $$\frac{\sec\frac12+1}{\tan\frac12}=\cot\frac14.$$ Therefore, $$T_2=\cot^{-1}(\cot\tfrac14)=\frac14,$$ since $\frac14\in(0,\pi)$. --- 3. **Compute the required value** The expression is $$T_1-T_2=(\pi-1)-\frac14=\pi-\frac54.$$ --- 4. **Match with the options** $$\pi-\frac54$$ corresponds to **Option C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.
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