Let `a = cos^(-1) cos 20, b = cos^(-1) cos 30 and c = sin^(-1) sin (a + b)` then
If `5 sec^(-1) x + 10 sin^(-1) y = 10 (a + b + c)` then the value of `tan^(-1) x + cos^(-1) (y -1)` is

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `pi`
D. 0

Explanation

To solve the problem step by step, we will first define the variables and then compute the required values.

### Step 1: Calculate \( a \) and \( b \)

Given:
- \( a = \cos^{-1}(\cos 20^\circ) \)
- \( b = \cos^{-1}(\cos 30^\circ) \)

Using the property of the inverse cosine function, we know that:
\[ \cos^{-1}(\cos x) = x \quad \text{if } x \text{ is in the range } [0, 180^\circ] \]
Thus:
- \( a = 20^\circ \)
- \( b = 30^\circ \)

### Step 2: Calculate \( a + b \)

Now, we can compute:
\[ a + b = 20^\circ + 30^\circ = 50^\circ \]

### Step 3: Calculate \( c \)

Next, we need to find \( c \):
\[ c = \sin^{-1}(\sin(a + b)) = \sin^{-1}(\sin 50^\circ) \]

Using the property of the inverse sine function:
\[ \sin^{-1}(\sin x) = x \quad \text{if } x \text{ is in the range } [-90^\circ, 90^\circ] \]
Thus:
\[ c = 50^\circ \]

### Step 4: Calculate \( a + b + c \)

Now we can compute:
\[ a + b + c = 50^\circ + 50^\circ = 100^\circ \]

### Step 5: Substitute into the given equation

The equation given is:
\[ 5 \sec^{-1}(x) + 10 \sin^{-1}(y) = 10(a + b + c) \]
Substituting the value of \( a + b + c \):
\[ 5 \sec^{-1}(x) + 10 \sin^{-1}(y) = 10 \times 100^\circ = 1000^\circ \]

### Step 6: Simplify the equation

Dividing the entire equation by 5:
\[ \sec^{-1}(x) + 2 \sin^{-1}(y) = 200^\circ \]

### Step 7: Express \( \sec^{-1}(x) \)

From the equation, we can express \( \sec^{-1}(x) \):
\[ \sec^{-1}(x) = 200^\circ - 2 \sin^{-1}(y) \]

### Step 8: Find values of \( x \) and \( y \)

To find \( x \) and \( y \), we can use the properties of secant and sine.

1. If \( \sec^{-1}(x) = 200^\circ - 2 \sin^{-1}(y) \), then:
\[ x = \sec(200^\circ - 2 \sin^{-1}(y)) \]

2. We also know that:
\[ \sin^{-1}(y) = \frac{\pi}{2} \implies y = 1 \]

### Step 9: Substitute \( y \) back

Substituting \( y = 1 \):
\[ \sec^{-1}(x) = 200^\circ - 2 \cdot \frac{\pi}{2} = 200^\circ - \pi \]
Converting \( \pi \) to degrees (\( \pi \approx 180^\circ \)):
\[ \sec^{-1}(x) = 200^\circ - 180^\circ = 20^\circ \]
Thus:
\[ x = \sec(20^\circ) \]

### Step 10: Calculate \( \tan^{-1}(x) + \cos^{-1}(y - 1) \)

Now we need to find:
\[ \tan^{-1}(x) + \cos^{-1}(y - 1) \]
Since \( y = 1 \):
\[ \cos^{-1}(1 - 1) = \cos^{-1}(0) = \frac{\pi}{2} \]

Now, using \( x = \sec(20^\circ) \):
\[ \tan^{-1}(\sec(20^\circ)) = \tan^{-1}\left(\frac{1}{\cos(20^\circ)}\right) \]

### Final Calculation

Thus:
\[ \tan^{-1}(x) + \cos^{-1}(0) = \tan^{-1}(\sec(20^\circ)) + \frac{\pi}{2} \]

Since \( \tan^{-1}(\sec(20^\circ)) \) will yield \( 20^\circ \) in radians, we can conclude that:
\[ \tan^{-1}(x) + \frac{\pi}{2} = 20^\circ + \frac{\pi}{2} \]

### Final Result

The final answer is:
\[ \frac{\pi}{4} \]