If the number of solutions of `sin^-1 x+|x|=1 cos^-1 x+|x|=1, tan^-1 x+|x|=1, cot^-1x+|x|=1, sec^-1 x+|x|=1 and cosec^-1 x+|x|=1` are `n_1,n_2,n_3,n_4,n_5,n_6` respectively, then the value of `n_1+n_2+n_3+n_4+n_5+n_6` is

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem, we need to analyze the equations involving the inverse trigonometric functions and the absolute value function. We will find the number of solutions for each equation one by one.

### Step 1: Analyze \( \sin^{-1} x + |x| = 1 \)

1. The domain of \( \sin^{-1} x \) is \( x \in [-1, 1] \).
2. Consider two cases for \( |x| \):
- Case 1: \( x \geq 0 \) → \( |x| = x \)
\[ \sin^{-1} x + x = 1 \]
- Case 2: \( x < 0 \) → \( |x| = -x \)
\[ \sin^{-1} x - x = 1 \]

**Finding solutions:**
- For Case 1, \( \sin^{-1} x + x = 1 \) can be solved graphically or numerically within the interval \( [0, 1] \).
- For Case 2, \( \sin^{-1} x - x = 1 \) is valid in the interval \( [-1, 0) \).

After solving both cases, we find that there are **2 solutions** for \( n_1 \).

### Step 2: Analyze \( \cos^{-1} x + |x| = 1 \)

1. The domain of \( \cos^{-1} x \) is \( x \in [-1, 1] \).
2. Again, consider two cases for \( |x| \):
- Case 1: \( x \geq 0 \) → \( |x| = x \)
\[ \cos^{-1} x + x = 1 \]
- Case 2: \( x < 0 \) → \( |x| = -x \)
\[ \cos^{-1} x - x = 1 \]

**Finding solutions:**
- For Case 1, \( \cos^{-1} x + x = 1 \) has **1 solution** in \( [0, 1] \).
- For Case 2, \( \cos^{-1} x - x = 1 \) has **1 solution** in \( [-1, 0) \).

Thus, \( n_2 = 2 \).

### Step 3: Analyze \( \tan^{-1} x + |x| = 1 \)

1. The function \( \tan^{-1} x \) is defined for all \( x \).
2. Again, consider two cases for \( |x| \):
- Case 1: \( x \geq 0 \) → \( |x| = x \)
\[ \tan^{-1} x + x = 1 \]
- Case 2: \( x < 0 \) → \( |x| = -x \)
\[ \tan^{-1} x - x = 1 \]

**Finding solutions:**
- Both cases can be solved graphically or numerically. Each case yields **1 solution**.

Thus, \( n_3 = 2 \).

### Step 4: Analyze \( \cot^{-1} x + |x| = 1 \)

1. The function \( \cot^{-1} x \) is defined for all \( x \).
2. Consider two cases for \( |x| \):
- Case 1: \( x \geq 0 \) → \( |x| = x \)
\[ \cot^{-1} x + x = 1 \]
- Case 2: \( x < 0 \) → \( |x| = -x \)
\[ \cot^{-1} x - x = 1 \]

**Finding solutions:**
- Each case yields **1 solution**.

Thus, \( n_4 = 2 \).

### Step 5: Analyze \( \sec^{-1} x + |x| = 1 \)

1. The domain of \( \sec^{-1} x \) is \( |x| \geq 1 \).
2. Consider two cases for \( |x| \):
- Case 1: \( x \geq 1 \) → \( |x| = x \)
\[ \sec^{-1} x + x = 1 \]
- Case 2: \( x < -1 \) → \( |x| = -x \)
\[ \sec^{-1} x - x = 1 \]

**Finding solutions:**
- Each case yields **1 solution**.

Thus, \( n_5 = 2 \).

### Step 6: Analyze \( \csc^{-1} x + |x| = 1 \)

1. The domain of \( \csc^{-1} x \) is \( |x| \geq 1 \).
2. Consider two cases for \( |x| \):
- Case 1: \( x \geq 1 \) → \( |x| = x \)
\[ \csc^{-1} x + x = 1 \]
- Case 2: \( x < -1 \) → \( |x| = -x \)
\[ \csc^{-1} x - x = 1 \]

**Finding solutions:**
- Each case yields **1 solution**.

Thus, \( n_6 = 2 \).

### Final Calculation

Now we sum the number of solutions:
\[ n_1 + n_2 + n_3 + n_4 + n_5 + n_6 = 2 + 2 + 2 + 2 + 2 + 2 = 12 \]

### Final Answer
The value of \( n_1 + n_2 + n_3 + n_4 + n_5 + n_6 \) is **12**.