Question 37 of 42advanced🔧 ApplyLong Answer4 marks

Prove that:

cot⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x]=x2,x∈(0,π4)\cot^{-1}\left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right] = \frac{x}{2}, \quad x \in \left(0, \frac{\pi}{4}\right)
Correct Answer

LHS:

Let

y=cot⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x].y = \cot^{-1}\left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right].

We simplify the terms inside the square roots using half-angle identities. From the context of Example 4, we use the identities:

1+sin⁡x=cos⁡2x2+sin⁡2x2+2sin⁡x2cos⁡x2=(cos⁡x2+sin⁡x2)21+\sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 1−sin⁡x=cos⁡2x2+sin⁡2x2−2sin⁡x2cos⁡x2=(cos⁡x2−sin⁡x2)21-\sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2

Since x∈(0,π/4)x \in (0, \pi/4), we have x2∈(0,π8)\frac{x}{2} \in (0, \frac{\pi}{8}). In this interval, cos⁡x2>sin⁡x2>0\cos\frac{x}{2} > \sin\frac{x}{2} > 0. Therefore, the square roots are:

1+sin⁡x=cos⁡x2+sin⁡x2\sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2} 1−sin⁡x=cos⁡x2−sin⁡x2\sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}

Substituting these into the expression:

y=cot⁡−1[(cos⁡x2+sin⁡x2)+(cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)−(cos⁡x2−sin⁡x2)]y = \cot^{-1}\left[\frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) + \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) - \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}\right]

Simplifying the numerator and denominator:

Numerator =2cos⁡x2= 2\cos\frac{x}{2}

Denominator =2sin⁡x2= 2\sin\frac{x}{2}

So,

y=cot⁡−1[2cos⁡x22sin⁡x2]=cot⁡−1(cot⁡x2).y = \cot^{-1}\left[\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right] = \cot^{-1}\left(\cot\frac{x}{2}\right).

Since x2\frac{x}{2} lies in the principal range of cot⁡−1\cot^{-1}, we have:

y=x2.y = \frac{x}{2}.

∴ Hence Proved.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 9 | (Page 14)
For More Understanding

Explanation

This question is solved by simplifying the trigonometric expression inside the inverse function. The key step involves rationalizing or simplifying 1±sin⁡x\sqrt{1 \pm \sin x} using half-angle formulas (1±sin⁡x=(cos⁡x2±sin⁡x2)21 \pm \sin x = \left(\cos\frac{x}{2} \pm \sin\frac{x}{2}\right)^2), a technique illustrated in Example 4 of the provided context.

The domain x∈(0,π/4)x \in (0, \pi/4) is crucial to determine the signs of the trigonometric functions, ensuring the square roots yield positive values and the inverse function is well-defined.

Solution Steps

  1. Step 1: Express 1+sin⁡x1+\sin x and 1−sin⁡x1-\sin x as perfect squares using the identities 1+sin⁡x=(cos⁡x2+sin⁡x2)21+\sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 and 1−sin⁡x=(cos⁡x2−sin⁡x2)21-\sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2.

  2. Step 2: Evaluate the square roots 1+sin⁡x\sqrt{1+\sin x} and 1−sin⁡x\sqrt{1-\sin x}. Since x∈(0,π/4)x \in (0, \pi/4), cos⁡x2>sin⁡x2\cos\frac{x}{2} > \sin\frac{x}{2}, so 1+sin⁡x=cos⁡x2+sin⁡x2\sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2} and 1−sin⁡x=cos⁡x2−sin⁡x2\sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}.

  3. Step 3: Substitute these values into the LHS expression and simplify the fraction to cos⁡x2sin⁡x2=cot⁡x2\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} = \cot\frac{x}{2}.

  4. Step 4: Apply the inverse cotangent function: cot⁡−1(cot⁡x2)=x2\cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}, which matches the RHS.

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