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Question: If $\sin^{-1}\sqrt{x^2+2x+1} + \sec^{-1}\sqrt{x^2+2x+1} = \frac{\pi}{2}, x \neq 0$, then the value o...

If sin⁡−1x2+2x+1+sec⁡−1x2+2x+1=π2,x≠0\sin^{-1}\sqrt{x^2+2x+1} + \sec^{-1}\sqrt{x^2+2x+1} = \frac{\pi}{2}, x \neq 0, then the value of 2sec⁡−1x2+sin⁡−1x2=2\sec^{-1}\frac{x}{2} + \sin^{-1}\frac{x}{2} =

A

3π2\frac{3\pi}{2}

Answer

3π2\frac{3\pi}{2}

Explanation

Solution

The given equation is sin⁡−1x2+2x+1+sec⁡−1x2+2x+1=π2\sin^{-1}\sqrt{x^2+2x+1} + \sec^{-1}\sqrt{x^2+2x+1} = \frac{\pi}{2}.

First, simplify the term inside the square root: x2+2x+1=(x+1)2=∣x+1∣\sqrt{x^2+2x+1} = \sqrt{(x+1)^2} = |x+1|.

Let y=∣x+1∣y = |x+1|. The equation becomes sin⁡−1y+sec⁡−1y=π2\sin^{-1}y + \sec^{-1}y = \frac{\pi}{2}.

For this equation to be defined, yy must be in the intersection of the domains of sin⁡−1\sin^{-1} and sec⁡−1\sec^{-1}.

The domain of sin⁡−1y\sin^{-1}y is [−1,1][-1, 1].

The domain of sec⁡−1y\sec^{-1}y is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty).

The intersection of these domains is [−1,1]∩((−∞,−1]∪[1,∞))={−1,1}[-1, 1] \cap ((-\infty, -1] \cup [1, \infty)) = \{-1, 1\}.

So, yy must be either −1-1 or 11.

Since y=∣x+1∣≥0y = |x+1| \ge 0, we must have y=1y=1.

Thus, ∣x+1∣=1|x+1| = 1.

This gives two possibilities: x+1=1x+1 = 1 or x+1=−1x+1 = -1.

If x+1=1x+1 = 1, then x=0x = 0.

If x+1=−1x+1 = -1, then x=−2x = -2.

The problem states that x≠0x \neq 0. Therefore, we must have x=−2x = -2.

Now we need to find the value of the expression 2sec⁡−1x2+sin⁡−1x22\sec^{-1}\frac{x}{2} + \sin^{-1}\frac{x}{2} for x=−2x=-2.

Substitute x=−2x=-2 into the expression:

2sec⁡−1−22+sin⁡−1−22=2sec⁡−1(−1)+sin⁡−1(−1)2\sec^{-1}\frac{-2}{2} + \sin^{-1}\frac{-2}{2} = 2\sec^{-1}(-1) + \sin^{-1}(-1).

We need to find the values of sec⁡−1(−1)\sec^{-1}(-1) and sin⁡−1(−1)\sin^{-1}(-1).

For sin⁡−1(−1)\sin^{-1}(-1): The principal value of sin⁡−1z\sin^{-1}z is in the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. We are looking for an angle θ∈[−π2,π2]\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] such that sin⁡θ=−1\sin\theta = -1. This angle is θ=−π2\theta = -\frac{\pi}{2}. So, sin⁡−1(−1)=−π2\sin^{-1}(-1) = -\frac{\pi}{2}.

For sec⁡−1(−1)\sec^{-1}(-1): The principal value of sec⁡−1z\sec^{-1}z is in the range [0,π]∖{π2}[0, \pi] \setminus \{\frac{\pi}{2}\}. We are looking for an angle ϕ∈[0,π]∖{π2}\phi \in [0, \pi] \setminus \{\frac{\pi}{2}\} such that sec⁡ϕ=−1\sec\phi = -1. Since sec⁡ϕ=1cos⁡ϕ\sec\phi = \frac{1}{\cos\phi}, we have 1cos⁡ϕ=−1\frac{1}{\cos\phi} = -1, which means cos⁡ϕ=−1\cos\phi = -1. The angle in the range [0,π][0, \pi] with cos⁡ϕ=−1\cos\phi = -1 is ϕ=π\phi = \pi. So, sec⁡−1(−1)=π\sec^{-1}(-1) = \pi.

Now substitute these values into the expression:

2sec⁡−1(−1)+sin⁡−1(−1)=2(π)+(−π2)=2π−π2=4π−π2=3π22\sec^{-1}(-1) + \sin^{-1}(-1) = 2(\pi) + (-\frac{\pi}{2}) = 2\pi - \frac{\pi}{2} = \frac{4\pi - \pi}{2} = \frac{3\pi}{2}.

The value of the expression is 3π2\frac{3\pi}{2}.