RD Sharma Solutions for Class 12 Maths Exercise 4.1 Question 2
Understanding the Question 🧐
This question asks us to find the principal value for several inverse secant functions. The principal value is the unique value that lies within the defined range of the inverse trigonometric function.
Find the principal value of the following functions:
(i) &&sec^{-1}(-\sqrt{2})&&
(ii) &&sec^{-1}(2)&&
(iii) &&sec^{-1}(\frac{2}{\sqrt{3}})&&
(iv) &&sec^{-1}(-2)&&
Our Expert’s Approach
To solve this, our experts first identify the core concept: finding the principal value of &&sec^{-1}(x)&&. The key is to remember its principal value branch, which is &&[0, \pi] – \{\frac{\pi}{2}\}&&. For negative inputs, we use the specific identity &&sec^{-1}(-x) = \pi – sec^{-1}(x)&&. This methodical approach ensures the answer is always within the correct range. This method is chosen for its clarity and reliability, reflecting years of teaching experience in providing rd sharma solutions.
Part (i): Find the principal value of &&sec^{-1}(-\sqrt{2})&& 📝
Step 1: Define the equation
Let &&y = sec^{-1}(-\sqrt{2})&&. By definition, this means &&sec(y) = -\sqrt{2}&&.
Step 2: Use the identity for negative inputs
The identity for the inverse secant function with a negative argument is &&sec^{-1}(-x) = \pi – sec^{-1}(x)&&. Applying this here:
&&y = \pi – sec^{-1}(\sqrt{2})&&
Step 3: Find the value of &&sec^{-1}(\sqrt{2})&&
We need to find an angle &&\theta&& in the range &&[0, \pi] – \{\frac{\pi}{2}\}&& such that &&sec(\theta) = \sqrt{2}&&. We know that &&sec(\frac{\pi}{4}) = \sqrt{2}&&. So, &&sec^{-1}(\sqrt{2}) = \frac{\pi}{4}&&.
Step 4: Calculate the final principal value
Substitute the value back into the equation from Step 2:
&&y = \pi – \frac{\pi}{4} = \frac{4\pi – \pi}{4} = \frac{3\pi}{4}&&
The principal value of &&sec^{-1}(-\sqrt{2})&& is &&\frac{3\pi}{4}&&.
Part (ii): Find the principal value of &&sec^{-1}(2)&& 📝
Step 1: Define the equation
Let &&y = sec^{-1}(2)&&. This implies &&sec(y) = 2&&.
Step 2: Identify the angle from standard values
We need to find the angle &&y&& in the principal value branch &&[0, \pi] – \{\frac{\pi}{2}\}&& whose secant is &&2&&. From our knowledge of trigonometry, we know that &&sec(\frac{\pi}{3}) = 2&&.
Step 3: Verify the range
The value &&\frac{\pi}{3}&& lies within the principal value range &&[0, \pi]&&.
Therefore, the principal value of &&sec^{-1}(2)&& is &&\frac{\pi}{3}&&.
Part (iii): Find the principal value of &&sec^{-1}(\frac{2}{\sqrt{3}})&& 📝
Step 1: Define the equation
Let &&y = sec^{-1}(\frac{2}{\sqrt{3}})&&. This means &&sec(y) = \frac{2}{\sqrt{3}}&&.
Step 2: Identify the angle from standard values
We look for an angle &&y&& in the range &&[0, \pi] – \{\frac{\pi}{2}\}&&. We know that &&sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}}&&.
Step 3: Verify the range
The angle &&\frac{\pi}{6}&& is clearly within the principal value branch.
Thus, the principal value of &&sec^{-1}(\frac{2}{\sqrt{3}})&& is &&\frac{\pi}{6}&&.
Part (iv): Find the principal value of &&sec^{-1}(-2)&& 📝
Step 1: Define the equation
Let &&y = sec^{-1}(-2)&&, which means &&sec(y) = -2&&.
Step 2: Use the identity for negative inputs
We apply the identity &&sec^{-1}(-x) = \pi – sec^{-1}(x)&&:
&&y = \pi – sec^{-1}(2)&&
Step 3: Find the value of &&sec^{-1}(2)&&
As we found in Part (ii), we know that &&sec^{-1}(2) = \frac{\pi}{3}&&.
Step 4: Calculate the final principal value
Substitute this value back into the equation:
&&y = \pi – \frac{\pi}{3} = \frac{3\pi – \pi}{3} = \frac{2\pi}{3}&&
The principal value of &&sec^{-1}(-2)&& is &&\frac{2\pi}{3}&&.
Conclusion and Key Points ✅
By applying the definition of the principal value branch of the inverse secant function and the specific identity for negative inputs, we have found the solutions:
- The principal value of &&sec^{-1}(-\sqrt{2})&& is &&\frac{3\pi}{4}&&.
- The principal value of &&sec^{-1}(2)&& is &&\frac{\pi}{3}&&.
- The principal value of &&sec^{-1}(\frac{2}{\sqrt{3}})&& is &&\frac{\pi}{6}&&.
- The principal value of &&sec^{-1}(-2)&& is &&\frac{2\pi}{3}&&.
- The principal value branch of &&sec^{-1}(x)&& is &&[0, \pi] – \{\frac{\pi}{2}\}&&.
- The identity &&sec^{-1}(-x) = \pi – sec^{-1}(x)&& is crucial for solving problems with negative arguments.
- Always ensure your final answer lies within the defined principal value range. An answer outside this range is incorrect.
Frequently Asked Questions (FAQ)
What is the principal value branch of &&sec^{-1}(x)&&?
The principal value branch for the inverse secant function, &&sec^{-1}(x)&&, is the interval &&[0, \pi]&& excluding &&\frac{\pi}{2}&&. This is written as &&[0, \pi] – \{\frac{\pi}{2}\}&&.
Why is &&\frac{\pi}{2}&& excluded from the range of &&sec^{-1}(x)&&?
The value &&\frac{\pi}{2}&& is excluded because &&sec(x) = \frac{1}{cos(x)}&&, and &&cos(\frac{\pi}{2}) = 0&&. Division by zero is undefined, so &&sec(\frac{\pi}{2})&& is undefined. Consequently, no value of &&x&& exists for which &&sec^{-1}(x)&& can be &&\frac{\pi}{2}&&, so it is removed from the range.
How do you find the principal value of &&sec^{-1}(x)&& for a negative input &&x&&?
For a negative input (where &&x > 0&&), you use the identity: &&sec^{-1}(-x) = \pi – sec^{-1}(x)&&. This formula helps map the negative input to an equivalent expression within the principal value range.
What is the relationship between &&sec^{-1}(x)&& and &&cos^{-1}(x)&&?
The relationship is &&sec^{-1}(x) = cos^{-1}(\frac{1}{x})&& for &&|x| \ge 1&&. This is derived from the fundamental relationship where &&sec(\theta) = \frac{1}{cos(\theta)}&&.
Is &&sec^{-1}(x)&& the same as &&(sec(x))^{-1}&&?
No, they are different. &&sec^{-1}(x)&& represents the inverse secant function (or arcsecant), which gives an angle. In contrast, &&(sec(x))^{-1}&& is the multiplicative inverse of &&sec(x)&&, which is equal to &&\frac{1}{sec(x)}&& or simply &&cos(x)&&.
Further Reading
For a deeper understanding of Inverse Trigonometric Functions, their properties, and graphs, we recommend visiting an authoritative resource like the Khan Academy article on Inverse Trigonometric Functions.
