Samacheer Kalvi Class 12 Maths Solutions Chapter 4 Inverse Trigonometric Functions Exercise 4.3

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.3

Chapter 4

Question 1. Find the domain of the following functions
(i) \( \tan^{-1} (\sqrt {9-x^2}) \)
(ii) \( \frac {1}{2} \tan^{-1} (1 – x^2) – \frac {π}{4} \)
Answer:
(i) Let \( f(x) = \tan^{-1}(\sqrt{9-x^2}) \).
The domain of \( \tan^{-1}(y) \) is \( (-\infty, \infty) \). So, for \( f(x) \) to be defined, the term inside the tangent inverse, \( \sqrt{9-x^2} \), must be a real number.
Also, for \( \sqrt{9-x^2} \) to be a real number, the expression under the square root must be non-negative.
So, \( 9-x^2 \geq 0 \)
\( \implies 9 \geq x^2 \)
\( \implies x^2 \leq 9 \)
\( \implies |x| \leq 3 \)
This means \( -3 \leq x \leq 3 \). Thus, the domain of \( f(x) \) is the interval \( [-3, 3] \). This ensures that the square root gives a real, non-negative value.
In simple words: For the function to work, the part under the square root must be zero or more. Also, the square root's answer must be a real number. This limits x to be between -3 and 3 (including these numbers).

(ii) Let \( g(x) = \frac{1}{2} \tan^{-1}(1-x^2) - \frac{\pi}{4} \).
The domain of \( \tan^{-1}(y) \) is \( (-\infty, \infty) \). This means the argument \( (1-x^2) \) can be any real number.
So, we need \( -\infty < 1-x^2 < \infty \).
This inequality is true for all real values of \( x \). Any real number \( x \) will make \( 1-x^2 \) a real number, which is within the domain of \( \tan^{-1} \).
Therefore, the domain of \( g(x) \) is \( \mathbb{R} \), which includes all real numbers. The other terms in the function, \( \frac{1}{2} \) and \( -\frac{\pi}{4} \), are constants and do not affect the domain.
In simple words: For this function, the number inside the \( \tan^{-1} \) can be any real number. Since \( 1-x^2 \) always gives a real number for any \( x \), the function works for all real numbers.

🎯 Exam Tip: When finding the domain of inverse trigonometric functions, always check the domain restrictions of both the inverse function itself and any nested functions (like square roots or fractions) that make up its argument.

Question 2. Find the value of
(i) \( \tan^{-1}(\tan\frac {5π}{4}) \)
(ii) \( \tan^{-1}(\tan(-\frac {π}{6})) \)
Answer:
(i) We need to find the value of \( \tan^{-1}(\tan\frac {5π}{4}) \).
The principal value branch of \( \tan^{-1}x \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
First, simplify \( \tan\frac{5\pi}{4} \):
\( \tan\frac{5\pi}{4} = \tan(\pi + \frac{\pi}{4}) \)
\( \implies \tan(\pi + \frac{\pi}{4}) = \tan(\frac{\pi}{4}) \) (since tan is positive in the third quadrant and has a period of \( \pi \))
Now, substitute this back into the expression:
\( \tan^{-1}(\tan\frac{5\pi}{4}) = \tan^{-1}(\tan\frac{\pi}{4}) \)
\( \implies \tan^{-1}(\tan\frac{\pi}{4}) = \frac{\pi}{4} \) (since \( \frac{\pi}{4} \) lies within the principal value branch \( (-\frac{\pi}{2}, \frac{\pi}{2}) \))
In simple words: First, change \( \tan(5\pi/4) \) to \( \tan(\pi/4) \) because \( 5\pi/4 \) is in the third quarter of the circle where tan is positive. Since \( \pi/4 \) is a normal angle, the answer is just \( \pi/4 \).

(ii) We need to find the value of \( \tan^{-1}(\tan(-\frac {π}{6})) \).
The principal value branch of \( \tan^{-1}x \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
In this case, the angle \( -\frac{\pi}{6} \) already lies within the principal value branch \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
Therefore, we can directly apply the property \( \tan^{-1}(\tan x) = x \) when \( x \) is in the principal value branch.
\( \tan^{-1}(\tan(-\frac{π}{6})) = -\frac{π}{6} \)
In simple words: The angle \( -\pi/6 \) is already in the special range where \( \tan^{-1}(\tan x) \) is simply \( x \). So, the answer is \( -\pi/6 \).

🎯 Exam Tip: Always check if the angle \( x \) in \( \tan^{-1}(\tan x) \) lies within the principal value branch of \( \tan^{-1}x \), which is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). If it doesn't, adjust the angle using trigonometric identities to bring it into this range.

Question 3. Find the value of the following:
(i) \( \tan(\tan^{-1}(\frac{7\pi}{4})) \)
(ii) \( \tan(\tan^{-1}(1947)) \)
(iii) \( \tan(\tan^{-1}(-0.2021)) \)
Answer:
We use the property that for any real number \( x \), \( \tan(\tan^{-1}x) = x \). This property holds because the domain of \( \tan^{-1}x \) is all real numbers, so any value \( x \) can be taken as an input.
(i) \( \tan(\tan^{-1}(\frac{7\pi}{4})) = \frac{7\pi}{4} \)
(ii) \( \tan(\tan^{-1}(1947)) = 1947 \)
(iii) \( \tan(\tan^{-1}(-0.2021)) = -0.2021 \)
In simple words: When you have \( \tan \) and \( \tan^{-1} \) next to each other, they cancel each other out. So, the answer is simply the number that was inside the \( \tan^{-1} \) function. This works for any real number.

🎯 Exam Tip: Remember the fundamental inverse trigonometric property: \( f(f^{-1}(x)) = x \) for \( x \) in the domain of \( f^{-1} \). For \( \tan(\tan^{-1}x) \), the domain of \( \tan^{-1}x \) is all real numbers, so this identity always applies.

Question 4. Find the value of
(i) \( \tan(\cos^{-1}(\frac {1}{2}) – \sin^{-1}(-\frac {1}{2})) \)
(ii) \( \sin(\tan^{-1}(\frac {1}{2}) – \cos^{-1}(\frac {4}{5})) \)
(iii) \( \cos(\sin^{-1} (\frac {4}{5}) – \tan^{-1} (\frac {3}{4})) \)
Answer:
(i) We need to find \( \tan(\cos^{-1}(\frac{1}{2}) – \sin^{-1}(-\frac{1}{2})) \).
First, evaluate the inverse trigonometric terms:
\( \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \) (since \( \cos(\frac{\pi}{3}) = \frac{1}{2} \) and \( \frac{\pi}{3} \in [0, \pi] \))
Using the property \( \sin^{-1}(-x) = -\sin^{-1}x \):
\( \sin^{-1}(-\frac{1}{2}) = -\sin^{-1}(\frac{1}{2}) \)
\( \implies -\sin^{-1}(\frac{1}{2}) = -\frac{\pi}{6} \) (since \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) and \( -\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \))
Now substitute these values back into the original expression:
\( \tan(\frac{\pi}{3} - (-\frac{\pi}{6})) = \tan(\frac{\pi}{3} + \frac{\pi}{6}) \)
\( \implies \tan(\frac{2\pi + \pi}{6}) = \tan(\frac{3\pi}{6}) \)
\( \implies \tan(\frac{\pi}{2}) \)
The value of \( \tan(\frac{\pi}{2}) \) is undefined, which is often represented as \( \infty \). A tangent of \( 90^\circ \) or \( \pi/2 \) approaches infinity.
In simple words: First, find the angles for \( \cos^{-1}(1/2) \) and \( \sin^{-1}(-1/2) \). Then, subtract them. Finally, find the \( \tan \) of the resulting angle. The result is undefined or infinity.

(ii) We need to find \( \sin(\tan^{-1}(\frac{1}{2}) – \cos^{-1}(\frac{4}{5})) \).
Let \( a = \tan^{-1}(\frac{1}{2}) \) and \( b = \cos^{-1}(\frac{4}{5}) \). We need to find \( \sin(a-b) \).
We know that \( \sin(a-b) = \sin a \cos b - \cos a \sin b \).
From \( a = \tan^{-1}(\frac{1}{2}) \), we can form a right-angled triangle. Opposite side = 1, Adjacent side = 2. Hypotenuse = \( \sqrt{1^2+2^2} = \sqrt{5} \).
So, \( \sin a = \frac{1}{\sqrt{5}} \) and \( \cos a = \frac{2}{\sqrt{5}} \). (Since \( \tan^{-1}(\frac{1}{2}) \) lies in the first quadrant, \( \sin a \) and \( \cos a \) are positive).
From \( b = \cos^{-1}(\frac{4}{5}) \), we have \( \cos b = \frac{4}{5} \). (Since \( \cos^{-1}(\frac{4}{5}) \) lies in the first quadrant, \( \sin b \) is positive).
We can find \( \sin b \) using \( \sin^2 b + \cos^2 b = 1 \):
\( \sin b = \sqrt{1 - \cos^2 b} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25-16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \).
Now, substitute these values into the formula for \( \sin(a-b) \):
\( \sin(a-b) = (\frac{1}{\sqrt{5}})(\frac{4}{5}) - (\frac{2}{\sqrt{5}})(\frac{3}{5}) \)
\( \implies \sin(a-b) = \frac{4}{5\sqrt{5}} - \frac{6}{5\sqrt{5}} \)
\( \implies \sin(a-b) = \frac{4-6}{5\sqrt{5}} = \frac{-2}{5\sqrt{5}} \)
In simple words: We want to find \( \sin(A-B) \). First, we convert the \( \tan^{-1} \) and \( \cos^{-1} \) parts into their sine and cosine values by drawing triangles. Then, we use the formula for \( \sin(A-B) \) to get the final answer.

(iii) We need to find \( \cos(\sin^{-1} (\frac{4}{5}) – \tan^{-1} (\frac{3}{4})) \).
Let \( a = \sin^{-1}(\frac{4}{5}) \) and \( b = \tan^{-1}(\frac{3}{4}) \). We need to find \( \cos(a-b) \).
We know that \( \cos(a-b) = \cos a \cos b + \sin a \sin b \).
From \( a = \sin^{-1}(\frac{4}{5}) \), we have \( \sin a = \frac{4}{5} \). (Since \( \sin^{-1}(\frac{4}{5}) \) lies in the first quadrant, \( \cos a \) is positive).
We can find \( \cos a \) using \( \sin^2 a + \cos^2 a = 1 \):
\( \cos a = \sqrt{1 - \sin^2 a} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \).
From \( b = \tan^{-1}(\frac{3}{4}) \), we can form a right-angled triangle. Opposite side = 3, Adjacent side = 4. Hypotenuse = \( \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5 \).
So, \( \sin b = \frac{3}{5} \) and \( \cos b = \frac{4}{5} \). (Since \( \tan^{-1}(\frac{3}{4}) \) lies in the first quadrant, \( \sin b \) and \( \cos b \) are positive).
Now, substitute these values into the formula for \( \cos(a-b) \):
\( \cos(a-b) = (\frac{3}{5})(\frac{4}{5}) + (\frac{4}{5})(\frac{3}{5}) \)
\( \implies \cos(a-b) = \frac{12}{25} + \frac{12}{25} \)
\( \implies \cos(a-b) = \frac{12+12}{25} = \frac{24}{25} \)
In simple words: We want to find \( \cos(A-B) \). First, we find the \( \sin \) and \( \cos \) values for each inverse trigonometric term using triangles. Then, we plug these values into the \( \cos(A-B) \) formula to find the final answer.

🎯 Exam Tip: When dealing with sums or differences of inverse trigonometric functions, convert each inverse function into basic trigonometric ratios (sine, cosine) using right-angled triangles and then apply the standard compound angle formulas like \( \sin(A \pm B) \) or \( \cos(A \pm B) \).