Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.11

Question 1. Find the principal value of
(i) \( \sin^{-1} (\frac{1}{\sqrt{2}}) \)
(ii) \( \cos^{-1} (\frac{\sqrt{3}}{2}) \)
(iii) \( \csc^{-1} (-1) \)
(iv) \( \sec^{-1} (-\sqrt{2}) \)
(v) \( \tan^{-1} (\sqrt{3}) \)
Answer:
(i) To find the principal value of \( \sin^{-1} (\frac{1}{\sqrt{2}}) \):
Let \( y = \sin^{-1} (\frac{1}{\sqrt{2}}) \). The principal branch of \( \sin^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
So, \( \sin y = \frac{1}{\sqrt{2}} \). We know that \( \sin (\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \).
Therefore, \( y = \frac{\pi}{4} \). This value lies within the principal range.
The principal value of \( \sin^{-1} (\frac{1}{\sqrt{2}}) \) is \( \frac{\pi}{4} \).
(ii) To find the principal value of \( \cos^{-1} (\frac{\sqrt{3}}{2}) \):
Let \( y = \cos^{-1} (\frac{\sqrt{3}}{2}) \). The principal branch of \( \cos^{-1} \) is \( [0, \pi] \).
So, \( \cos y = \frac{\sqrt{3}}{2} \). We know that \( \cos (\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).
Therefore, \( y = \frac{\pi}{6} \). This value lies within the principal range.
The principal value of \( \cos^{-1} (\frac{\sqrt{3}}{2}) \) is \( \frac{\pi}{6} \).
(iii) To find the principal value of \( \csc^{-1} (-1) \):
Let \( y = \csc^{-1} (-1) \). The principal branch of \( \csc^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\} \).
So, \( \csc y = -1 \). This means \( \frac{1}{\sin y} = -1 \), which gives \( \sin y = -1 \).
We know that \( \sin (-\frac{\pi}{2}) = -1 \).
Therefore, \( y = -\frac{\pi}{2} \). This value lies within the principal range.
The principal value of \( \csc^{-1} (-1) \) is \( -\frac{\pi}{2} \).
(iv) To find the principal value of \( \sec^{-1} (-\sqrt{2}) \):
Let \( y = \sec^{-1} (-\sqrt{2}) \). The principal branch of \( \sec^{-1} \) is \( [0, \pi] - \{\frac{\pi}{2}\} \).
So, \( \sec y = -\sqrt{2} \). This means \( \frac{1}{\cos y} = -\sqrt{2} \), which gives \( \cos y = -\frac{1}{\sqrt{2}} \).
We know that \( \cos (\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \). Since \( \cos y \) is negative, \( y \) must be in the second quadrant.
Therefore, \( y = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \). This value lies within the principal range.
The principal value of \( \sec^{-1} (-\sqrt{2}) \) is \( \frac{3\pi}{4} \).
(v) To find the principal value of \( \tan^{-1} (\sqrt{3}) \):
Let \( y = \tan^{-1} (\sqrt{3}) \). The principal branch of \( \tan^{-1} \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
So, \( \tan y = \sqrt{3} \). We know that \( \tan (\frac{\pi}{3}) = \sqrt{3} \).
Therefore, \( y = \frac{\pi}{3} \). This value lies within the principal range.
The principal value of \( \tan^{-1} (\sqrt{3}) \) is \( \frac{\pi}{3} \).
In simple words: For each inverse trigonometric function, we find the angle whose value matches the given number, making sure the angle is within the specific allowed range for that function. This allowed range is called the principal branch.

🎯 Exam Tip: Always remember the principal value branches for each inverse trigonometric function as they define the unique output. For negative values, pay attention to the quadrant defined by the principal range.

Question 2. A man standing directly opposite to one side of a road of width x meter views a circular shaped traffic green signal of diameter 'a' meter on the other side of the road. The bottom of the green signal Is 'b' meter height from the horizontal level of viewer's eye. If \( \alpha \) denotes the angle subtended by the diameter of the green signal at the viewer's eye, then prove that \( \alpha = \tan^{-1} (\frac{a+b}{x}) - \tan^{-1} (\frac{b}{x}) \)
Answer:
Let's set up the situation. We have a man on one side of a road with width \( x \). On the other side is a traffic light.
The green signal has a diameter of \( a \) meters.
The bottom of the signal is \( b \) meters high from the man's eye level.
This means the top of the signal is at a height of \( (a+b) \) meters from the man's eye level.

Imagine a right-angled triangle formed by the man's eye, the road (base \( x \)), and the vertical line to the top of the signal (height \( a+b \)).
Let \( \phi \) be the angle the top of the signal makes at the man's eye.
We can use the tangent function: \( \tan \phi = \frac{\text{Opposite}}{\text{Adjacent}} \)
\( \tan \phi = \frac{a+b}{x} \)
\( \implies \phi = \tan^{-1} (\frac{a+b}{x}) \)

Next, consider another right-angled triangle formed by the man's eye, the road (base \( x \)), and the vertical line to the bottom of the signal (height \( b \)).
Let \( \theta \) be the angle the bottom of the signal makes at the man's eye.
Using the tangent function again:
\( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \)
\( \tan \theta = \frac{b}{x} \)
\( \implies \theta = \tan^{-1} (\frac{b}{x}) \)

The angle subtended by the diameter of the green signal at the viewer's eye, \( \alpha \), is the difference between these two angles.
It is the angle from the top of the signal to the bottom of the signal, as seen by the man.
\( \alpha = \phi - \theta \)
Now, substitute the expressions for \( \phi \) and \( \theta \):
\( \implies \alpha = \tan^{-1} (\frac{a+b}{x}) - \tan^{-1} (\frac{b}{x}) \)
This proves the given statement. The difference between the angles to the top and bottom of the signal gives the angle it covers in the man's view.
In simple words: We find the angle from the man's eye to the very top of the signal. Then we find the angle from his eye to the very bottom of the signal. The difference between these two angles is the size of the signal as seen by him.

🎯 Exam Tip: For problems involving angles of elevation or depression, always draw a clear diagram to visualize the right-angled triangles and label the known and unknown values. The tangent function is often very useful for relating angles to opposite and adjacent sides.