Samacheer Kalvi Class 12 Maths Solutions Chapter 4 Inverse Trigonometric Functions Exercise 4.4

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Question 1. Find the principle value of
(i) \( \sec^{-1} \left(\frac{2}{\sqrt{3}}\right) \)
(ii) \( \cot^{-1} (\sqrt{3}) \)
(iii) \( \operatorname{cosec}^{-1} (-\sqrt{2}) \)
Answer:
(i) Let \( \sec^{-1} \left(\frac{2}{\sqrt{3}}\right) = \theta \).
So, \( \sec \theta = \frac{2}{\sqrt{3}} \). This means \( \cos \theta = \frac{\sqrt{3}}{2} \).
Since \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), we have \( \theta = \frac{\pi}{6} \). The principal value branch for \( \sec^{-1} \) is \( [0, \pi] - \{\frac{\pi}{2}\} \).
Thus, the principal value is \( \frac{\pi}{6} \).
(ii) Let \( \cot^{-1} (\sqrt{3}) = \theta \).
So, \( \cot \theta = \sqrt{3} \).
Since \( \cot \frac{\pi}{6} = \sqrt{3} \), we have \( \theta = \frac{\pi}{6} \). The principal value branch for \( \cot^{-1} \) is \( (0, \pi) \).
Thus, the principal value is \( \frac{\pi}{6} \).
(iii) Let \( \operatorname{cosec}^{-1} (-\sqrt{2}) = \theta \).
So, \( \operatorname{cosec} \theta = -\sqrt{2} \). This means \( \sin \theta = -\frac{1}{\sqrt{2}} \).
We know \( \operatorname{cosec} \frac{\pi}{4} = \sqrt{2} \), so \( -\operatorname{cosec} \frac{\pi}{4} = -\sqrt{2} \).
So, \( \operatorname{cosec} \theta = \operatorname{cosec} (-\frac{\pi}{4}) \). The principal value branch for \( \operatorname{cosec}^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\} \).
Thus, the principal value is \( -\frac{\pi}{4} \).
In simple words: To find the principal value, we figure out which angle in the main range of the inverse trigonometric function gives the given value. For example, for \( \sec^{-1} \), we look for an angle between 0 and \( \pi \) (not including \( \frac{\pi}{2} \)) that matches the secant value.

🎯 Exam Tip: Always remember the principal value branches for each inverse trigonometric function to pick the correct angle.

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Question 2. Find the value
(i) \( \tan^{-1} (\sqrt{3}) – \sec^{-1}(-2) \)
(ii) \( \sin^{-1}(-1) + \cos^{-1}\left(\frac{1}{2}\right) + \cot^{-1}(2) \)
(iii) \( \cot^{-1}(1) + \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) – \sec^{-1}(-\sqrt{2}) \)
Answer:
(i) For \( \tan^{-1}(\sqrt{3}) \):
Let \( x = \tan^{-1}(\sqrt{3}) \).
Then \( \tan x = \sqrt{3} \).
Since \( \tan \frac{\pi}{3} = \sqrt{3} \), we get \( x = \frac{\pi}{3} \).
For \( \sec^{-1}(-2) \):
Let \( y = \sec^{-1}(-2) \).
Then \( \sec y = -2 \).
We know \( \sec \frac{\pi}{3} = 2 \). So, \( \sec y = -\sec \frac{\pi}{3} \).
This can be written as \( \sec y = \sec (\pi - \frac{\pi}{3}) \) because \( \sec(\pi - \alpha) = -\sec \alpha \).
So, \( \sec y = \sec \left(\frac{3\pi - \pi}{3}\right) = \sec \left(\frac{2\pi}{3}\right) \).
Therefore, \( y = \frac{2\pi}{3} \).
Now, we calculate the expression:
\( \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} \)
\( = \frac{\pi - 2\pi}{3} = -\frac{\pi}{3} \).
(ii) For \( \sin^{-1}(-1) \):
Let \( x = \sin^{-1}(-1) \).
Then \( \sin x = -1 \).
Since \( \sin (-\frac{\pi}{2}) = -1 \), we get \( x = -\frac{\pi}{2} \).
For \( \cos^{-1}\left(\frac{1}{2}\right) \):
Let \( y = \cos^{-1}\left(\frac{1}{2}\right) \).
Then \( \cos y = \frac{1}{2} \).
Since \( \cos \frac{\pi}{3} = \frac{1}{2} \), we get \( y = \frac{\pi}{3} \).
For \( \cot^{-1}(2) \):
Let \( z = \cot^{-1}(2) \).
Then \( \cot z = 2 \). This value cannot be expressed as a standard angle, so it remains as \( \cot^{-1}(2) \).
Now, we calculate the expression:
\( \sin^{-1}(-1) + \cos^{-1}\left(\frac{1}{2}\right) + \cot^{-1}(2) = -\frac{\pi}{2} + \frac{\pi}{3} + \cot^{-1}(2) \)
\( = \frac{-3\pi + 2\pi}{6} + \cot^{-1}(2) \)
\( = -\frac{\pi}{6} + \cot^{-1}(2) \) or \( \cot^{-1}(2) - \frac{\pi}{6} \).
(iii) For \( \cot^{-1}(1) \):
Let \( x = \cot^{-1}(1) \).
Then \( \cot x = 1 \).
Since \( \cot \frac{\pi}{4} = 1 \), we get \( x = \frac{\pi}{4} \).
For \( \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \):
Let \( y = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \).
Then \( \sin y = -\frac{\sqrt{3}}{2} \).
Since \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \), we have \( \sin y = -\sin \frac{\pi}{3} \).
This can be written as \( \sin y = \sin (-\frac{\pi}{3}) \) because \( \sin(-\alpha) = -\sin \alpha \).
So, \( y = -\frac{\pi}{3} \).
For \( \sec^{-1}(-\sqrt{2}) \):
Let \( z = \sec^{-1}(-\sqrt{2}) \).
Then \( \sec z = -\sqrt{2} \).
We know \( \sec \frac{\pi}{4} = \sqrt{2} \). So, \( \sec z = -\sec \frac{\pi}{4} \).
This can be written as \( \sec z = \sec (\pi - \frac{\pi}{4}) \).
So, \( \sec z = \sec \left(\frac{4\pi - \pi}{4}\right) = \sec \left(\frac{3\pi}{4}\right) \).
Therefore, \( z = \frac{3\pi}{4} \).
Now, we calculate the expression:
\( \cot^{-1}(1) + \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) - \sec^{-1}(-\sqrt{2}) = \frac{\pi}{4} + \left(-\frac{\pi}{3}\right) - \frac{3\pi}{4} \)
\( = \frac{\pi}{4} - \frac{\pi}{3} - \frac{3\pi}{4} \)
To combine these, find a common denominator, which is 12:
\( = \frac{3\pi}{12} - \frac{4\pi}{12} - \frac{9\pi}{12} \)
\( = \frac{3\pi - 4\pi - 9\pi}{12} \)
\( = \frac{-10\pi}{12} \)
\( = -\frac{5\pi}{6} \).
In simple words: To find the value of these expressions, we first find the principal value for each inverse trigonometric term separately. Then, we add and subtract these values according to the signs in the question. Sometimes, we need to use trigonometric identities to find the correct angle in the principal range.

🎯 Exam Tip: When dealing with negative values for inverse trigonometric functions, remember the specific quadrants and ranges for each function to ensure you pick the correct principal value. For example, \( \sec^{-1}(-x) = \pi - \sec^{-1}(x) \).