GSEB Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2.1

Find the Principal Values of the Following:

Question 1. \( \sin^{-1}\left(-\frac{1}{2}\right) \)
Answer: Assume \( \sin^{-1}\left(-\frac{1}{2}\right) = y \). This means that \( \sin y = -\frac{1}{2} \). We know \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), so \( \sin y = -\sin\left(\frac{\pi}{6}\right) \). Using the property \( -\sin x = \sin(-x) \), we get \( \sin y = \sin\left(-\frac{\pi}{6}\right) \). The main range for the principal value branch of \( \sin^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Therefore, the principal value for \( \sin^{-1}\left(-\frac{1}{2}\right) \) becomes \( -\frac{\pi}{6} \).
In simple words: To find the main value, let the expression equal 'y'. Then, find the angle 'y' whose sine is \( -\frac{1}{2} \), making sure 'y' is between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

Exam Tip: Always remember the principal value range for each inverse trigonometric function. For \( \sin^{-1}x \), the range is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

Question 2. \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \)
Answer: Assume \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = y \). This implies that \( \cos y = -\frac{\sqrt{3}}{2} \). We know \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), so \( \cos y = -\cos\left(\frac{\pi}{6}\right) \). Using the identity \( -\cos x = \cos(\pi - x) \), we get \( \cos y = \cos\left(\pi - \frac{\pi}{6}\right) \), which simplifies to \( \cos y = \cos\left(\frac{5\pi}{6}\right) \). The primary range for the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \). Therefore, the principal value for \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \) is \( \frac{5\pi}{6} \).
In simple words: Set the expression to 'y'. Find the angle 'y' whose cosine is \( -\frac{\sqrt{3}}{2} \). Ensure 'y' falls between 0 and \( \pi \), using the \( \pi - x \) rule for negative values.

Exam Tip: For inverse cosine of negative values, use the identity \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \) to stay within the \( [0, \pi] \) range.

Question 3. \( \csc^{-1}(2) \)
Answer: Assume \( \csc^{-1}(2) = y \). This implies that \( \csc y = 2 \). We know that \( \csc\left(\frac{\pi}{6}\right) = 2 \). The primary range for the principal value branch of \( \csc^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), excluding \( \{0\} \). Therefore, the principal value for \( \csc^{-1}(2) \) is \( \frac{\pi}{6} \).
In simple words: Let the expression be 'y'. Find the angle 'y' for which the cosecant is 2. Make sure 'y' is within the main allowed range for cosecant, which is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) but not zero.

Exam Tip: Remember that \( \csc^{-1}(x) = \sin^{-1}\left(\frac{1}{x}\right) \). You can convert to inverse sine and use its range if you forget the cosecant range directly.

Question 4. \( \tan^{-1}(-\sqrt{3}) \)
Answer: Assume \( \tan^{-1}(-\sqrt{3}) = y \). This means that \( \tan y = -\sqrt{3} \). We know \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), so \( \tan y = -\tan\left(\frac{\pi}{3}\right) \). Using the identity \( -\tan x = \tan(-x) \), we get \( \tan y = \tan\left(-\frac{\pi}{3}\right) \). The primary range for the principal value branch of \( \tan^{-1} \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Therefore, the principal value for \( \tan^{-1}(-\sqrt{3}) \) is \( -\frac{\pi}{3} \).
In simple words: Let the expression be 'y'. Find the angle 'y' whose tangent is \( -\sqrt{3} \). Make sure 'y' is in the open interval from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

Exam Tip: For inverse tangent of negative values, simply apply the negative sign to the angle, as \( \tan^{-1}(-x) = -\tan^{-1}(x) \) and the range is symmetric about zero.

Question 5. \( \cos^{-1}\left(-\frac{1}{2}\right) \)
Answer: Assume \( \cos^{-1}\left(-\frac{1}{2}\right) = y \). This implies that \( \cos y = -\frac{1}{2} \). We know \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), so \( \cos y = -\cos\left(\frac{\pi}{3}\right) \). Using the identity \( -\cos x = \cos(\pi - x) \), we get \( \cos y = \cos\left(\pi - \frac{\pi}{3}\right) \), which simplifies to \( \cos y = \cos\left(\frac{2\pi}{3}\right) \). The primary range for the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \). Therefore, the principal value for \( \cos^{-1}\left(-\frac{1}{2}\right) \) is \( \frac{2\pi}{3} \).
In simple words: Set the expression to 'y'. Find the angle 'y' whose cosine is \( -\frac{1}{2} \). Ensure 'y' falls between 0 and \( \pi \), using the \( \pi - x \) rule for negative values.

Exam Tip: Remember that for negative values of \( \cos^{-1} \), the angle will always be in the second quadrant to remain within the principal range \( [0, \pi] \).

Question 6. \( \tan^{-1}(-1) \)
Answer: Assume \( \tan^{-1}(-1) = y \). This indicates that \( \tan y = -1 \). We know \( \tan\left(\frac{\pi}{4}\right) = 1 \), so \( \tan y = -\tan\left(\frac{\pi}{4}\right) \). Using the property \( -\tan x = \tan(-x) \), we find \( \tan y = \tan\left(-\frac{\pi}{4}\right) \). The primary range for the principal value branch of \( \tan^{-1} \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Therefore, the principal value for \( \tan^{-1}(-1) \) is \( -\frac{\pi}{4} \).
In simple words: Let the expression be 'y'. Find the angle 'y' whose tangent is -1. Make sure 'y' is in the open interval from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

Exam Tip: Similar to \( \sin^{-1} \), for negative inputs to \( \tan^{-1} \), the output angle will be negative and within the range \( (-\frac{\pi}{2}, 0) \).

Question 7. \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \)
Answer: Assume \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) = y \). This means that \( \sec y = -\frac{2}{\sqrt{3}} \). We know \( \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \), so \( \sec y = -\sec\left(\frac{\pi}{6}\right) \). Using the identity \( -\sec x = \sec(\pi - x) \), we get \( \sec y = \sec\left(\pi - \frac{\pi}{6}\right) \), which simplifies to \( \sec y = \sec\left(\frac{5\pi}{6}\right) \). The principal value branch for \( \sec^{-1} \) is \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \). Therefore, the principal value for \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \) is \( \frac{5\pi}{6} \).
In simple words: Let the expression be 'y'. Find the angle 'y' whose secant is \( -\frac{2}{\sqrt{3}} \). Make sure 'y' is between 0 and \( \pi \), but not \( \frac{\pi}{2} \). Use the \( \pi - x \) rule for negative values.

Exam Tip: Convert to inverse cosine if you find inverse secant challenging: \( \sec^{-1}(-x) = \cos^{-1}\left(-\frac{1}{x}\right) \). Then apply the \( \pi - \cos^{-1}\left(\frac{1}{x}\right) \) formula.

Question 8. \( \cot^{-1}(\sqrt{3}) \)
Answer: Assume \( \cot^{-1}(\sqrt{3}) = y \). This implies that \( \cot y = \sqrt{3} \). We know \( \cot\left(\frac{\pi}{6}\right) = \sqrt{3} \). The primary range for the principal value branch of \( \cot^{-1} \) is \( (0, \pi) \). Therefore, the principal value for \( \cot^{-1}(\sqrt{3}) \) is \( \frac{\pi}{6} \).
In simple words: Let the expression be 'y'. Find the angle 'y' whose cotangent is \( \sqrt{3} \). Make sure 'y' is in the open interval from 0 to \( \pi \).

Exam Tip: The principal value range for \( \cot^{-1}x \) is \( (0, \pi) \), which is different from \( \tan^{-1}x \). Do not confuse the two ranges.

Question 9. \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
Answer: Assume \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = y \). This means that \( \cos y = -\frac{1}{\sqrt{2}} \). We know \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), so \( \cos y = -\cos\left(\frac{\pi}{4}\right) \). Using the identity \( -\cos x = \cos(\pi - x) \), we get \( \cos y = \cos\left(\pi - \frac{\pi}{4}\right) \), which simplifies to \( \cos y = \cos\left(\frac{3\pi}{4}\right) \). The primary range for the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \). Therefore, the principal value for \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \) is \( \frac{3\pi}{4} \).
In simple words: Set the expression to 'y'. Find the angle 'y' whose cosine is \( -\frac{1}{\sqrt{2}} \). Ensure 'y' falls between 0 and \( \pi \), using the \( \pi - x \) rule for negative values.

Exam Tip: Remember \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \). This helps ensure your angle is in the correct principal range for negative inputs.

Question 10. \( \csc^{-1}(-\sqrt{2}) \)
Answer: Assume \( \csc^{-1}(-\sqrt{2}) = y \). This implies that \( \csc y = -\sqrt{2} \). We know \( \csc\left(\frac{\pi}{4}\right) = \sqrt{2} \), so \( \csc y = -\csc\left(\frac{\pi}{4}\right) \). Using the identity \( -\csc x = \csc(-x) \), we get \( \csc y = \csc\left(-\frac{\pi}{4}\right) \). The primary range for the principal value branch of \( \csc^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), excluding \( \{0\} \). Therefore, the principal value for \( \csc^{-1}(-\sqrt{2}) \) is \( -\frac{\pi}{4} \).
In simple words: Let the expression be 'y'. Find the angle 'y' whose cosecant is \( -\sqrt{2} \). Make sure 'y' is within the main allowed range for cosecant, which is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) but not zero.

Exam Tip: The principal value range for \( \csc^{-1}x \) is identical to \( \sin^{-1}x \) except for excluding 0. This means for negative inputs, the angle will be negative.

Question 11. \( \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) \)
Answer:Let's find the principal value for each term:
1. For \( \tan^{-1}(1) \):
Assume \( x = \tan^{-1}(1) \). So, \( \tan x = 1 \). The principal value branch for \( \tan^{-1} \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Thus, \( x = \frac{\pi}{4} \).
2. For \( \cos^{-1}\left(-\frac{1}{2}\right) \):
Assume \( y = \cos^{-1}\left(-\frac{1}{2}\right) \). So, \( \cos y = -\frac{1}{2} \). We know \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), so \( \cos y = -\cos\left(\frac{\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) \). The principal value branch for \( \cos^{-1} \) is \( [0, \pi] \). Thus, \( y = \frac{2\pi}{3} \).
3. For \( \sin^{-1}\left(-\frac{1}{2}\right) \):
Assume \( z = \sin^{-1}\left(-\frac{1}{2}\right) \). So, \( \sin z = -\frac{1}{2} \). We know \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), so \( \sin z = -\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right) \). The principal value branch for \( \sin^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Thus, \( z = -\frac{\pi}{6} \).
Now, add these values together:
\( \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right) \)
\( = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} \)
\( = \frac{(3+8-2)\pi}{12} \)
\( = \frac{9\pi}{12} \)
\( = \frac{3\pi}{4} \)
In simple words: Calculate the main value for each of the three inverse trigonometric parts separately. Then, add all these main values together to get the final answer.

Exam Tip: When evaluating expressions with multiple inverse trigonometric functions, always find the principal value for each individual function first, then perform the arithmetic operation.

Question 12. \( \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) \)
Answer:Let's find the principal value for each term:
1. For \( \cos^{-1}\left(\frac{1}{2}\right) \):
Assume \( x = \cos^{-1}\left(\frac{1}{2}\right) \). So, \( \cos x = \frac{1}{2} \). The principal value branch for \( \cos^{-1} \) is \( [0, \pi] \). Thus, \( x = \frac{\pi}{3} \).
2. For \( \sin^{-1}\left(\frac{1}{2}\right) \):
Assume \( y = \sin^{-1}\left(\frac{1}{2}\right) \). So, \( \sin y = \frac{1}{2} \). The principal value branch for \( \sin^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Thus, \( y = \frac{\pi}{6} \).
Now, substitute these values into the expression:
\( \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2\left(\frac{\pi}{6}\right) \)
\( = \frac{\pi}{3} + \frac{2\pi}{6} \)
\( = \frac{\pi}{3} + \frac{\pi}{3} \)
\( = \frac{2\pi}{3} \)
In simple words: First, find the main angle for inverse cosine of \( \frac{1}{2} \), and the main angle for inverse sine of \( \frac{1}{2} \). Then, multiply the inverse sine result by 2 and add it to the inverse cosine result.

Exam Tip: Remember to perform multiplication before addition according to the order of operations when evaluating such expressions.

Question 13. If \( \sin^{-1} x = y \), then
(a) \( 0 \leq y \leq \pi \)
(b) \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
(c) \( 0 < y < \pi \)
(d) \( -\frac{\pi}{2} < y < \frac{\pi}{2} \)
Answer: (b) \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
In simple words: The range of possible principal values for the inverse sine function, \( \sin^{-1}x \), is always from negative \( \frac{\pi}{2} \) to positive \( \frac{\pi}{2} \), including both endpoints. This means the angle 'y' must fall within this specific closed interval.

Exam Tip: Thoroughly memorize the principal value ranges for all inverse trigonometric functions. This is a fundamental concept for solving related problems.

Question 14. \( \tan^{-1}\sqrt{3} - \sec^{-1}(-2) \) is equal to
(A) \( \pi \)
(B) \( -\frac{\pi}{3} \)
(C) \( \frac{\pi}{3} \)
(D) \( \frac{2\pi}{3} \)
Answer: (B) \( -\frac{\pi}{3} \)
In simple words: Calculate the main value of \( \tan^{-1}(\sqrt{3}) \). Then, calculate the main value of \( \sec^{-1}(-2) \). Finally, subtract the second result from the first result to get the answer.

Exam Tip: For \( \sec^{-1}(-x) \), use the property \( \pi - \sec^{-1}(x) \) to correctly find its principal value within the range \( [0, \pi] - \{\frac{\pi}{2}\} \).