If `x in(0,pi/2)` ,then `(sin^(-1)(cosx)+cos^(-1)(sinx))/(tan^(-1)(cotx)+cot^(-1)(tanx))=`

Options

A. `pi//4`
B. `pi//3`
C. `pi//2`
D. 1

Answer

To solve the expression

\[ \frac{\sin^{-1}(\cos x) + \cos^{-1}(\sin x)}{\tan^{-1}(\cot x) + \cot^{-1}(\tan x)} \]

for \( x \in (0, \frac{\pi}{2}) \), we can use some trigonometric identities.

### Step 1: Simplifying the numerator

We know the identity:

\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \]

for any \( y \) in the range \([0, 1]\). Here, we can apply this identity with \( y = \sin x \) and \( y = \cos x \):

\[ \sin^{-1}(\cos x) + \cos^{-1}(\sin x) = \frac{\pi}{2} \]

### Step 2: Simplifying the denominator

Similarly, we can use the identity:

\[ \tan^{-1}(y) + \cot^{-1}(y) = \frac{\pi}{2} \]

for any \( y > 0 \). Here, we can apply this identity with \( y = \tan x \):

\[ \tan^{-1}(\cot x) + \cot^{-1}(\tan x) = \frac{\pi}{2} \]

### Step 3: Putting it all together

Now substituting these results back into our original expression, we have:

\[ \frac{\frac{\pi}{2}}{\frac{\pi}{2}} = 1 \]

Thus, the value of the expression is:

\[ \boxed{1} \]