1.Principal Value: Inverse Trig Class 12

Principal Value: Inverse Trig Class 12

In this article,”Principal Value: Inverse Trig Class 12″, we will learn to find out the principal value of inverse trigonometric functions.

Also Read This Article :- Parametric Function Derivative Example:Solved Example and Formula

2.Key Points Summary of Inverse Trig Class 12

Principal Value: Inverse Trig Class 12

(1.)Basic Condition of Inverse Function
(i)The inverse (f^{-1}) of any function f is only possible if it is ‘one-one’ and ‘onto’ (i.e. bijective).
(ii)Trigonometric functions in the natural domain are not bijective,so they cannot be directly inversed.
(2.)Restrictions on Domain & Principal Value Branch
(i)The domains of trigonometric functions are restricted to extract the inverse so that they become one-one and onto.
(ii)On the basis of limited domains,many branches are formed,the main branch of which is called ‘Principal Value Branch’.
(3.)Inverse Trigonometric Functions:
(i)Domain and Range (Principal Value Branch)
(ii)Below is the standard reference table for Principal Value
Inverse Trigonometric Functions: Domain and Range (Principal Value Branch)
Below is the standard reference table for Principal Value Branches:
\begin{array}{|l|l|l|} \hline \textbf{Inverse Function} & \textbf{Domain} & \textbf{Range (PVB)} \\ \hline \sin^{-1}(x) & [-1, 1] & \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ \hline \cos^{-1} x & [-1, 1] & [0, \pi] \\ \hline \tan^{-1} x & \mathbb{R} & \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ \hline \cot^{-1} x & \mathbb{R} & (0, \pi) \\ \hline \sec^{-1}(x) & \mathbb{R} - (-1, 1) & (0, \pi] - \left\{\frac{\pi}{2}\right\} \\ \hline \operatorname{cosec}^{-1}(x) & \mathbb{R} - (-1, 1) & \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\} \\ \hline \end{array}
(4.)Graph transformation (Graph Highlights)
(i)The graph of y = f^{-1}(x) is obtained by interchanging the X and Y axes in the graph of y=f(x).
(ii)This graph is a mirror image of the original function along the line y=x.
(5.)Key Properties
\sin\left(\sin^{-1}(x)\right) = x (where x \in [-1, 1])
\sin^{-1}(\sin x) = x (where x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right])

Also Read This Article:- Inverse trigonometric functions

3.Examples Based on Principal Value: Inverse Trig Class 12

Example: 1. \sin^{-1}\left(-\frac{1}{2}\right)
Solution: \sin^{-1}\left(-\frac{1}{2}\right)
Let \sin^{-1}\left(-\frac{1}{2}\right) = x
\Rightarrow \sin x = -\frac{1}{2}
\Rightarrow \sin x = \sin\left(-\frac{\pi}{6}\right)
x = -\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence principal value of \sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}
Example: 2. \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
Solution: \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
Let \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = x
\Rightarrow \cos x = \frac{\sqrt{3}}{2}
\Rightarrow \cos x = \cos\left(\frac{\pi}{6}\right)
x = \frac{\pi}{6} \in [0, \pi]
Hence Principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) is \frac{\pi}{6}
Example: 3. \operatorname{cosec}^{-1}(2)
Solution: \operatorname{cosec}^{-1}(2)
Let \operatorname{cosec}^{-1}(2) = x
\Rightarrow \operatorname{cosec} x = 2
\Rightarrow \operatorname{cosec} x = \operatorname{cosec}\left(\frac{\pi}{6}\right)
\Rightarrow x = \frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence principal value of \operatorname{cosec}^{-1}(2) is \frac{\pi}{6}
Example: 4. \tan^{-1}(-\sqrt{3})
Solution: \tan^{-1}(-\sqrt{3})
Let \tan^{-1}(-\sqrt{3}) = x
\Rightarrow \tan x = -\sqrt{3}
\Rightarrow \tan x = -\tan\left(\frac{\pi}{3}\right)
\Rightarrow \tan x = \tan\left(-\frac{\pi}{3}\right)
\Rightarrow x = -\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
Hence principal value of \tan^{-1}(-\sqrt{3}) is -\frac{\pi}{3}
Example: 5. \cos^{-1}\left(-\frac{1}{2}\right)
Solution: \cos^{-1}\left(-\frac{1}{2}\right)
Let \cos^{-1}\left(-\frac{1}{2}\right) = x
\Rightarrow \cos x = -\frac{1}{2}
\Rightarrow \cos x = -\cos\left(\frac{\pi}{3}\right)
= \cos\left(\pi - \frac{\pi}{3}\right)
\Rightarrow \cos x = \cos\left(\frac{2\pi}{3}\right)
x = \frac{2\pi}{3}
Hence principal value of
\cos^{-1}\left(-\frac{1}{2}\right) is \frac{2\pi}{3}
Example: 6. \tan^{-1}(-1)
Solution: \tan^{-1}(-1)
Let \tan^{-1}(-1) = x
\Rightarrow \tan x = -1
\Rightarrow \tan x = -\tan\left(\frac{\pi}{4}\right)
\Rightarrow \tan x = \tan\left(-\frac{\pi}{4}\right)
x = -\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
Hence principal value of
\tan^{-1}(-1) is -\frac{\pi}{4}

Principal Value: Inverse Trig Class 12

Example: 7. \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)
Solution: \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)
Let \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = x
\Rightarrow \sec x = \frac{2}{\sqrt{3}}
\Rightarrow \sec x = \sec\left(\frac{\pi}{6}\right)
x = \frac{\pi}{6} \in [0, \pi]
Hence principal value of
\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) is \frac{\pi}{6}
Example: 8. \cot^{-1}(\sqrt{3})
Solution: \cot^{-1}(\sqrt{3})
Let \cot^{-1}(\sqrt{3}) = x
\Rightarrow \cot x = \sqrt{3}
\Rightarrow \cot x = \cot\left(\frac{\pi}{6}\right)
x = \frac{\pi}{6} \in (0, \pi]
Hence principal value of
\cot^{-1}(\sqrt{3}) is \frac{\pi}{6}
Example: 9. \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)
Solution: \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)
Let \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = x
\Rightarrow \cos x = -\frac{1}{\sqrt{2}}
\Rightarrow \cos x = -\cos\left(\frac{\pi}{4}\right)
\Rightarrow \cos x = \cos\left(\pi - \frac{\pi}{4}\right)
\Rightarrow \cos x = \cos\left(\frac{3\pi}{4}\right)
x = \frac{3\pi}{4} \in [0, \pi]
Hence principal value of
\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) is \frac{3\pi}{4}
Example: 10. \operatorname{cosec}^{-1}(-\sqrt{2})
Solution: \operatorname{cosec}^{-1}(-\sqrt{2})
Let \operatorname{cosec}^{-1}(-\sqrt{2}) = x
\Rightarrow \operatorname{cosec} x = -\sqrt{2}
\Rightarrow \operatorname{cosec} x = -\operatorname{cosec} \frac{\pi}{4}
\Rightarrow \operatorname{cosec} x = \operatorname{cosec}\left(-\frac{\pi}{4}\right)
x = -\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence principal value of
\operatorname{cosec}^{-1}(-\sqrt{2}) is \left(-\frac{\pi}{4}\right)
Find the values of the following:
Example: 11. \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)
Solution: \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)
Principal value of
\tan^{-1}(1) = \frac{\pi}{4}, \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3},
\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}
\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)
= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}
= \frac{3\pi + 8\pi - 2\pi}{12}
= \frac{9\pi}{12} = \frac{3\pi}{4}
Example: 12. \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)
Solution: \cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2}
Principal value of \cos^{-1}\frac{1}{2} = \frac{\pi}{3}
\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}
Hence \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 \times \frac{\pi}{6}
= \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}
Example: 13. If \sin^{-1} x = y, then
(A) 0 \le y \le \pi (B) -\frac{\pi}{2} \le y \le \frac{\pi}{2}
(C) 0 < y < \pi (D) -\frac{\pi}{2} < y < \frac{\pi}{2}
Solution: \sin^{-1} x = y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
Hence -\frac{\pi}{2} \le y \le \frac{\pi}{2}
Alternate (B) is true.
Example: 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}
Solution: \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)
= \frac{\pi}{3} - \frac{2\pi}{3} = \frac{4\pi - 9\pi}{12} = -\frac{5\pi}{12}
evaluating \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} and \sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3},
so \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}
= -\frac{\pi}{3}
Alternate (B) is true
By solving the above examples,one can understand the Principal Value: Inverse Trig Class 12.

Also Read This Article:- Types of Functions in Class 12

4.Practice Questions of Principal Value: Inverse Trig Class 12 for Students

Principal Value: Inverse Trig Class 12

Find the principal value of the following
(1.) \sec^{-1}(\sqrt{2}) (2.) \operatorname{cosec}^{-1}(-1)
(3.) \sec^{-1}(-\sqrt{2})
Answers: (1.) \frac{\pi}{4} (2.) -\frac{\pi}{2} (3.) \frac{3\pi}{4}
By solving the above questions,you can understand the Principal Value: Inverse Trig Class 12 well because the concept is well understood when you solve it practically.
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5.Frequentaly Asked Questions Related to Principal Value: Inverse Trig Class 12

6.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”331 का  \frac{3}{8} आसन्न मान क्या होगा?
(What is the adjacent value of \frac{3}{8} of 331?)
*पिछली प्रश्नोत्तरी का उत्तर* 82
दी गई संख्या श्रृंखला का क्रम इस प्रकार है: [82]
70 \quad 68 \quad 74 \quad 68 \quad 78 \quad ? \quad 68

अतः प्रश्नचिन्ह (?) के स्थान पर संख्या 82 आएगी।
*Previous Quiz Solution*  
The order of the given series is as follows:  82

Hence, the question mark (?) will be replaced by the number 82.
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*