Inverse Trigonometry PYQ | JEE Main 2025 | Cot⁻¹ Simplification Trick

 

❓ Question

Evaluate

$$\cot^{-1}\!\left(\frac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right) - \cot^{-1}\!\left(\frac{\sqrt{1+\tan^2\left(\frac12\right)}+1}{\tan\left(\frac12\right)}\right).$$

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✍️ Solution

Let

$$A=\cot^{-1}\!\left(\frac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right),$$

and

$$B=\cot^{-1}\!\left(\frac{\sqrt{1+\tan^2\left(\frac12\right)}+1}{\tan\left(\frac12\right)}\right).$$

The required value is

$$A-B.$$

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Step 1: Evaluate $A$

Using

$$1+\tan^2\theta=\sec^2\theta,$$

we get

$$\sqrt{1+\tan^2(2)} = |\sec2|.$$

Since

$$2\text{ rad}\in\left(\frac{\pi}{2},\pi\right),$$
$$\sec2<0$$

so

$$|\sec2|=-\sec2.$$

Hence,

$$A = \cot^{-1}\!\left(\frac{-\sec2-1}{\tan2}\right).$$

Now,

$$\sec2=\frac1{\cos2}, \qquad \tan2=\frac{\sin2}{\cos2},$$

therefore,

$$\frac{-\sec2-1}{\tan2} = -\frac{1+\cos2}{\sin2}.$$

Using the identities

$$1+\cos2=2\cos^2(1),$$

and

$$\sin2=2\sin1\cos1,$$

we obtain

$$-\frac{1+\cos2}{\sin2} = -\cot1.$$

Thus,

$$A=\cot^{-1}(-\cot1).$$

Using

$$\cot^{-1}(-x)=\pi-\cot^{-1}(x),$$

we get

$$A=\pi-1.$$

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Step 2: Evaluate $B$

Again,

$$1+\tan^2\theta=\sec^2\theta.$$

Since

$$\frac12\text{ rad}\in\left(0,\frac{\pi}{2}\right),$$
$$\sec\!\left(\frac12\right)>0,$$

so

$$\sqrt{1+\tan^2\left(\frac12\right)} = \sec\!\left(\frac12\right).$$

Hence,

$$B = \cot^{-1}\!\left( \frac{\sec\left(\frac12\right)+1} {\tan\left(\frac12\right)} \right).$$

Now,

$$\frac{\sec\theta+1}{\tan\theta} = \frac{1+\cos\theta}{\sin\theta} = \cot\frac{\theta}{2}.$$

Taking

$$\theta=\frac12,$$
$$\frac{\sec\left(\frac12\right)+1} {\tan\left(\frac12\right)} = \cot\left(\frac14\right).$$

Therefore,

$$B = \cot^{-1}\!\left(\cot\frac14\right) = \frac14.$$

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Step 3: Required Value

$$A-B = (\pi-1)-\frac14.$$

Therefore,

$$A-B = \pi-\frac54.$$

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✅ Final Answer

$$\boxed{\pi-\frac54}$$