Question

\( \sin^{-1}(-\cos 2) + \cos^{-1}(\sin 3) + \tan^{-1}(\cot 5) = \)

• A. 7
• B. 5
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

When simplifying expressions like \( f^{-1}(g(x)) \), first use co-function identities to make \(g\) the same as \(f\). Then, ensure the argument of the inner function (e.g., \( y \) in \( f^{-1}(f(y)) \)) is within the principal value range of \( f^{-1} \). If not, adjust it by adding or subtracting multiples of the function's period (e.g., \(2\pi\) for sin/cos, \(\pi\) for tan).

Answer 1

The Correct Option is C

We evaluate each term separately. Note that the angles 2, 3, and 5 are in radians.
Term 1: \( \sin^{-1}(-\cos 2) \). Using \( \sin^{-1}(-x) = -\sin^{-1}(x) \), this is \( -\sin^{-1}(\cos 2) \).
Using the identity \( \cos x = \sin(\frac{\pi}{2} - x) \), we get \( -\sin^{-1}(\sin(\frac{\pi}{2} - 2)) \).
Since \( \frac{\pi}{2} - 2 \approx 1.57 - 2 = -0.43 \) is in the range of \( \sin^{-1} \), which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we have \( -(\frac{\pi}{2} - 2) = 2 - \frac{\pi}{2} \).
Term 2: \( \cos^{-1}(\sin 3) \). Using the identity \( \sin x = \cos(\frac{\pi}{2} - x) \), we get \( \cos^{-1}(\cos(\frac{\pi}{2} - 3)) \).
The range of \( \cos^{-1} \) is \( [0, \pi] \). The value \( \frac{\pi}{2} - 3 \approx 1.57 - 3 = -1.43 \) is not in this range.
We use the property \( \cos(y) = \cos(-y) \). Let's try \( \sin x = \cos(x - \frac{\pi}{2}) \). So \( \cos^{-1}(\cos(3 - \frac{\pi}{2})) \).
The value \( 3 - \frac{\pi}{2} \approx 3 - 1.57 = 1.43 \) is in the range \( [0, \pi] \). So, Term 2 is \( 3 - \frac{\pi}{2} \).
Term 3: \( \tan^{-1}(\cot 5) \). Using \( \cot x = \tan(\frac{\pi}{2} - x) \), we get \( \tan^{-1}(\tan(\frac{\pi}{2} - 5)) \).
The range of \( \tan^{-1} \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). The value \( \frac{\pi}{2} - 5 \approx 1.57 - 5 = -3.43 \) is not in this range.
We use the property that \( \tan(y) = \tan(y + n\pi) \). We need to find an integer n such that \( \frac{\pi}{2} - 5 + n\pi \) is in the range.
For \( n=2 \), \( \frac{\pi}{2} - 5 + 2\pi = \frac{5\pi}{2} - 5 \approx 7.85 - 5 = 2.85 \), which is outside.
For \( n=1 \), \( \frac{\pi}{2} - 5 + \pi = \frac{3\pi}{2} - 5 \approx 4.71 - 5 = -0.29 \), which is inside the range.
So, Term 3 is \( \frac{3\pi}{2} - 5 \).
Summing all three terms: \( (2 - \frac{\pi}{2}) + (3 - \frac{\pi}{2}) + (\frac{3\pi}{2} - 5) \).
\( = (2+3-5) + (-\frac{\pi}{2} - \frac{\pi}{2} + \frac{3\pi}{2}) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \).