Question 36 of 42intermediate🔧 ApplyLong Answer4 marks

Prove that:

tan⁡−1x=12cos⁡−1(1−x1+x),x∈[0,1]\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right), \quad x \in [0, 1]
Correct Answer

Let tan⁡−1x=θ\tan^{-1}\sqrt{x} = \theta. Then x=tan⁡θ\sqrt{x} = \tan \theta, so x=tan⁡2θx = \tan^2 \theta.

Now,

1−x1+x=1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\frac{1-x}{1+x} = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta

Therefore,

cos⁡−1(1−x1+x)=cos⁡−1(cos⁡2θ)\cos^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}(\cos 2\theta)

Since x∈[0,1]x \in [0, 1], we have x∈[0,1]\sqrt{x} \in [0, 1], so θ=tan⁡−1x∈[0,π/4]\theta = \tan^{-1}\sqrt{x} \in [0, \pi/4].

Hence 2θ∈[0,π/2]2\theta \in [0, \pi/2], and cos⁡−1(cos⁡2θ)=2θ\cos^{-1}(\cos 2\theta) = 2\theta.

Thus,

cos⁡−1(1−x1+x)=2θ=2⋅tan⁡−1x\cos^{-1}\left(\frac{1-x}{1+x}\right) = 2\theta = 2 \cdot \tan^{-1}\sqrt{x}

Therefore,

tan⁡−1x=12cos⁡−1(1−x1+x)\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 8 | (Page 14)
For More Understanding

Explanation

This question tests the substitution method for proving inverse trigonometric identities. The key insight is to substitute x=tan⁡2θx = \tan^2 \theta, which transforms the algebraic expression into a trigonometric one.

The formula cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} is a standard double-angle identity.

The domain restriction x∈[0,1]x \in [0, 1] ensures θ∈[0,π/4]\theta \in [0, \pi/4], making 2θ∈[0,π/2]2\theta \in [0, \pi/2], which justifies cos⁡−1(cos⁡2θ)=2θ\cos^{-1}(\cos 2\theta) = 2\theta without any adjustment.

Solution Steps

  1. Step 1: Let tan⁡−1x=θ\tan^{-1}\sqrt{x} = \theta, so x=tan⁡θ\sqrt{x} = \tan \theta and x=tan⁡2θx = \tan^2 \theta

  2. Step 2: Express 1−x1+x=1−tan⁡2θ1+tan⁡2θ=cos⁡2θ\dfrac{1-x}{1+x} = \dfrac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta

  3. Step 3: Apply cos⁡−1\cos^{-1} to get cos⁡−1(1−x1+x)=cos⁡−1(cos⁡2θ)\cos^{-1}\left(\dfrac{1-x}{1+x}\right) = \cos^{-1}(\cos 2\theta)

  4. Step 4: Use domain x∈[0,1]x \in [0,1] to show θ∈[0,π/4]\theta \in [0, \pi/4], so 2θ∈[0,π/2]2\theta \in [0, \pi/2]

  5. Step 5: Conclude cos⁡−1(cos⁡2θ)=2θ\cos^{-1}(\cos 2\theta) = 2\theta, giving the required result

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