Question 38 of 42advanced🔧 ApplyLong Answer4 marks

Prove that:

tan⁡−1[1+x−1−x1+x+1−x]=π4−12cos⁡−1x,−12≤x≤1\tan^{-1}\left[\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right] = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x, \quad -\frac{1}{\sqrt{2}} \leq x \leq 1

[Hint: Put x=cos⁡2θx = \cos 2\theta]

Correct Answer

LHS =tan⁡−1[1+x−1−x1+x+1−x]= \tan^{-1}\left[\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right]

Let x=cos⁡2θx = \cos 2\theta (as given in hint)

Then 1+x=1+cos⁡2θ=2cos⁡2θ=2cos⁡θ\sqrt{1+x} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta

And 1−x=1−cos⁡2θ=2sin⁡2θ=2sin⁡θ\sqrt{1-x} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta

Substituting in LHS:

LHS=tan⁡−1[2cos⁡θ−2sin⁡θ2cos⁡θ+2sin⁡θ]LHS = \tan^{-1}\left[\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right] =tan⁡−1[cos⁡θ−sin⁡θcos⁡θ+sin⁡θ]= \tan^{-1}\left[\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\right]

Dividing numerator and denominator by cos⁡θ\cos\theta:

=tan⁡−1[1−tan⁡θ1+tan⁡θ]= \tan^{-1}\left[\frac{1 - \tan\theta}{1 + \tan\theta}\right] =tan⁡−1[tan⁡(π4−θ)]= \tan^{-1}\left[\tan\left(\frac{\pi}{4} - \theta\right)\right]

Since for the given domain, (π4−θ)\left(\frac{\pi}{4} - \theta\right) lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), we have:

=π4−θ= \frac{\pi}{4} - \theta

Finding θ\theta in terms of xx:

Since x=cos⁡2θx = \cos 2\theta, we have cos⁡−1x=2θ\cos^{-1}x = 2\theta

Therefore, θ=12cos⁡−1x\theta = \frac{1}{2}\cos^{-1}x

Final Result:

LHS=π4−θ=π4−12cos⁡−1x=RHSLHS = \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x = RHS

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 10 | (Page 14)
For More Understanding

Explanation

This question from Miscellaneous Exercise Q10 uses substitution method. The key insight is recognizing that 1+cos⁡2θ\sqrt{1+\cos 2\theta} and 1−cos⁡2θ\sqrt{1-\cos 2\theta} simplify to 2cos⁡θ\sqrt{2}\cos\theta and 2sin⁡θ\sqrt{2}\sin\theta respectively using half-angle identities.

The domain −12≤x≤1-\frac{1}{\sqrt{2}} \leq x \leq 1 ensures that when x=cos⁡2θx = \cos 2\theta, θ\theta ranges from 00 to 3π8\frac{3\pi}{8}, keeping both cos⁡θ\cos\theta and sin⁡θ\sin\theta positive, and ensuring (π4−θ)\left(\frac{\pi}{4} - \theta\right) stays within the principal range of tan⁡−1\tan^{-1}.

Solution Steps

  1. Step 1: Substitute x=cos⁡2θx = \cos 2\theta as suggested in the hint

  2. Step 2: Simplify 1+x=1+cos⁡2θ=2cos⁡θ\sqrt{1+x} = \sqrt{1+\cos 2\theta} = \sqrt{2}\cos\theta using identity 1+cos⁡2θ=2cos⁡2θ1+\cos 2\theta = 2\cos^2\theta

  3. Step 3: Simplify 1−x=1−cos⁡2θ=2sin⁡θ\sqrt{1-x} = \sqrt{1-\cos 2\theta} = \sqrt{2}\sin\theta using identity 1−cos⁡2θ=2sin⁡2θ1-\cos 2\theta = 2\sin^2\theta

  4. Step 4: Substitute and simplify the fraction to cos⁡θ−sin⁡θcos⁡θ+sin⁡θ\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}

  5. Step 5: Divide by cos⁡θ\cos\theta to get 1−tan⁡θ1+tan⁡θ=tan⁡(π4−θ)\frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)

  6. Step 6: Apply tan⁡−1\tan^{-1} to get π4−θ\frac{\pi}{4} - \theta

  7. Step 7: Express θ=12cos⁡−1x\theta = \frac{1}{2}\cos^{-1}x since x=cos⁡2θx = \cos 2\theta

  8. Step 8: Conclude LHS=π4−12cos⁡−1x=RHSLHS = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x = RHS

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