Inverse Trigonometric Functions: Chapter 2 Links
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In this exercise, we will study the important properties of inverse trigonometric functions 2.2 and learn how to simplify expressions involving them. Questions are mainly based on identities such as \sin(\sin^{-1}x)=x, \cos(\cos^{-1}x)=x, \tan(\tan^{-1}x)=x etc. and related principal value concepts.
We will also use standard trigonometric identities like double-angle and triple-angle formulae to transform complicated inverse trigonometric expressions into their simplest forms. Careful attention is required while working with domains and principal values of inverse trigonometric functions.
Before solving the questions, make sure you are comfortable with the domains, ranges and principal values of all inverse trigonometric functions along with basic trigonometric identities.
Key Concepts
1. Basic Inverse Trigonometric Identities
Inverse trigonometric functions are defined to reverse the action of trigonometric functions within restricted domains.
Key identities include:
\sin(\sin^{-1}x)=x,\quad \cos(\cos^{-1}x)=x,\quad \tan(\tan^{-1}x)=x
These hold only when x lies in the respective domains of the inverse functions.
2. Principal Value Concept
Inverse trigonometric functions return only one value, called the principal value.
This is done by restricting the domain of trigonometric functions so that they become one-one.
For example, \sin^{-1}x always gives values in \left[-\frac{\pi}{2},\frac{\pi}{2}\right].
3. Principal Value Branches (Range)
- \sin^{-1}x:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
- \cos^{-1}x:[0,\pi]
- \tan^{-1}x:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
- \cot^{-1}x:(0,\pi)
- \sec^{-1}x:[0,\pi]-\left\{\frac{\pi}{2}\right\}
- \csc^{-1}x:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}
4. Domain Restrictions of Trigonometric Functions
Trigonometric functions are not one-one over their natural domains due to periodicity.
Therefore, they are restricted to specific intervals to define inverse functions properly.
For example, sine repeats values such that \sin 0=\sin \pi=0, so it is restricted to make it invertible.
5. Key Working Rule for Problems
While solving problems, always follow this:
1. Convert inverse trig function into its principal angle (PVB)
2. Replace and simplify using standard angles
3. Apply trigonometric identities if needed
4. Check final angle lies within correct range
Let us now solve all the NCERT questions step by step in this exercise of Inverse Trigonometric Functions 2.2.
Question 1: Inverse Trigonometric Functions 2.2
1. Prove that
3\sin^{-1}x=\sin^{-1}(3x-4x^3),\quad x\in\left[-\frac12,\frac12\right]
Solution
In such questions involving inverse trigonometric functions, we usually start by making a suitable substitution. For this, you should be comfortable with standard trigonometric identities and formulae.
Now, looking at the expression 3x-4x^3 on the RHS, the first thing that should come to mind is the triple-angle identity:
\sin3\theta=3\sin\theta-4\sin^3\theta
So, let
x=\sin\theta
Then,
\theta=\sin^{-1}x
Substituting x=\sin\theta in the RHS, we get
\sin^{-1}(3x-4x^3)=\sin^{-1}(3\sin\theta-4\sin^3\theta)
Using the identity
\sin3\theta=3\sin\theta-4\sin^3\theta
we get
\sin^{-1}(3x-4x^3)=\sin^{-1}(\sin3\theta)
Now,
\sin^{-1}(\sin y)=y
only when
y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
So, we must check whether 3\theta lies in this interval or not.
Since
x\in\left[-\frac12,\frac12\right]
therefore,
\theta=\sin^{-1}x\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right]
Multiplying throughout by 3,
3\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Hence,
\sin^{-1}(\sin3\theta)=3\theta
Therefore,
\sin^{-1}(3x-4x^3)=3\theta
But \theta=\sin^{-1}x
Thus we have,
\sin^{-1}(3x-4x^3)=3\sin^{-1}x
Hence proved.
Question 2: Inverse Trigonometric Functions 2.2
2. Prove that
3\cos^{-1}x=\cos^{-1}(4x^3-3x),\quad x\in\left[\frac12,1\right]
Solution
In such questions involving inverse trigonometric functions, we usually begin with a suitable substitution. So, one should be familiar with standard trigonometric identities and formulae.
Now, looking at the expression 4x^3-3x on the RHS, the first thing that should come to mind is the following triple-angle identity:
\cos3\theta=4\cos^3\theta-3\cos\theta
So, let
x=\cos\theta
Then,
\theta=\cos^{-1}x
Substituting x=\cos\theta in the RHS, we get
\cos^{-1}(4x^3-3x)=\cos^{-1}(4\cos^3\theta-3\cos\theta)
Using the identity
\cos3\theta=4\cos^3\theta-3\cos\theta
we get
\cos^{-1}(4x^3-3x)=\cos^{-1}(\cos3\theta)
Now,
\cos^{-1}(\cos y)=y
only when
y\in[0,\pi]
So, we must check whether 3\theta lies in this interval or not.
Since
x\in\left[\frac12,1\right]
therefore,
\theta=\cos^{-1}x\in\left[0,\frac{\pi}{3}\right]
Multiplying throughout by 3,
3\theta\in[0,\pi]
Hence,
\cos^{-1}(\cos3\theta)=3\theta
Therefore,
\cos^{-1}(4x^3-3x)=3\theta
But \theta=\cos^{-1}x
Thus we get,
\cos^{-1}(4x^3-3x)=3\cos^{-1}x
Hence proved.
You may already be following Maths Better for NCERT Solutions for the topics like Matrices, Determinants and Relations and Functions. Similarly, this exercise on Inverse Trigonometric Functions for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills.
Now, let’s move on to the next question of Inverse Trigonometric Functions 2.2.
Question 3: Inverse Trigo by Substitution
3. Write the following function in the simplest form :
\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right),\quad x\ne 0
Solution
In questions like these, the main idea is to observe the expression carefully and think of a suitable trigonometric substitution. Usually, the expression inside the bracket can be connected to some standard trigonometric identity.
Here, since we have terms involving \sqrt{1+x^2}, the substitution
x=\tan\theta
naturally comes to mind because
1+\tan^2\theta=\sec^2\theta
Let
x=\tan\theta
Then,
\theta=\tan^{-1}x
and
\sqrt{1+x^2}=\sqrt{1+\tan^2\theta}=\sec\theta
Substituting these in the given expression, we get
\tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right)
Now, writing in terms of sine and cosine,
\frac{\sec\theta-1}{\tan\theta}=\frac{\frac1{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}
=\frac{1-\cos\theta}{\sin\theta}
Now, using half-angle identities, we have
1-\cos\theta=2\sin^2\frac{\theta}{2}
and
\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}
Therefore,
\frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}
=\tan\frac{\theta}{2}
we get
\frac{\sec\theta-1}{\tan\theta}=\tan\frac{\theta}{2}
Therefore,
\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}\left(\tan\frac{\theta}{2}\right)
Now, since \theta=\tan^{-1}x, therefore
\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Hence,
\frac{\theta}{2}\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)
which lies inside the principal value branch of \tan^{-1}x. Therefore,
\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}
Thus,
\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{\theta}{2}
But \theta=\tan^{-1}x
Hence,
\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac12\tan^{-1}x
Question 4: Simplifing the Expression
4. Simplify :
\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),\quad 0 < x < \pi
Solution
In questions like these, we carefully observe the expression inside the inverse trigonometric function and try to connect it with a standard trigonometric identity.
Here, the expression
\sqrt{\frac{1-\cos x}{1+\cos x}}
suggests the half-angle identity involving \tan\frac{x}{2}.
Using the identities
1-\cos x=2\sin^2\frac{x}{2}
and
1+\cos x=2\cos^2\frac{x}{2}
we get
\sqrt{\frac{1-\cos x}{1+\cos x}}=\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}
=\sqrt{\tan^2\frac{x}{2}}
=\left|\tan\frac{x}{2}\right|
Now, since
0 < x < \pi
therefore,
0 < \frac{x}{2} < \frac{\pi}{2}
and in this interval,
\tan\frac{x}{2} > 0
Hence,
\left|\tan\frac{x}{2}\right|=\tan\frac{x}{2}
Therefore, the given expression becomes
\tan^{-1}\left(\tan\frac{x}{2}\right)
Now,
\tan^{-1}(\tan\theta)=\theta
only when
\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Since
0 < \frac{x}{2} < \frac{\pi}{2}
therefore, \frac{x}{2} lies in the principal value branch of \tan^{-1}x.
Hence,
\tan^{-1}\left(\tan\frac{x}{2}\right)=\frac{x}{2}
Therefore, the simplified form is
\boxed{\frac{x}{2}}
Question 5: Inverse Trigonometric Functions Ex. 2.2
5. Simplify :
\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right),\quad -\frac{\pi}{4} < x < \frac{3\pi}{4}
Solution
In such questions, we try to manipulate the expression inside the inverse trigonometric function so that it resembles a standard trigonometric identity.
Here, both numerator and denominator contain \sin x and \cos x. So, the first thing that should come to mind is dividing throughout by \cos x to convert the expression into terms of \tan x.
Thus,
\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}
Now, this resembles the standard identity
\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
Taking A=\frac{\pi}{4} and B=x, we get
\tan\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}
Therefore, the given expression becomes
\tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right)
Now,
\tan^{-1}(\tan\theta)=\theta
only when
\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Since
-\frac{\pi}{4} < x < \frac{3\pi}{4}
Multiply throughout by -1 , we get
\frac{\pi}{4} > -x > -\frac{3\pi}{4}
or -\frac{3\pi}{4} < -x < \frac{\pi}{4}
Adding \frac{\pi}{4} throughout, we get
-\frac{\pi}{2} < \frac{\pi}{4}-x < \frac{\pi}{2}
Thus, \frac{\pi}{4}-x lies in the principal value branch of \tan^{-1}x.
Hence,
\tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right)=\frac{\pi}{4}-x
Therefore, the simplified form is
\boxed{\frac{\pi}{4}-x}
While solving inverse trigonometric function questions, always remember that many angles may satisfy a trigonometric equation, but the final answer must belong to the principal value branch of the given inverse trigonometric function.
Question 6: Inverse Trigonometric Functions 2.2
6. Simplify :
\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right),\quad |x| < a
Solution
In such questions, the idea is to look at the expression carefully and think of a suitable trigonometric substitution.
Here, the expression contains
\sqrt{a^2-x^2}
which naturally suggests the substitution
x=a\sin\theta
because
1-\sin^2\theta=\cos^2\theta
Let
x=a\sin\theta
Then,
\theta=\sin^{-1}\left(\frac{x}{a}\right)
and
\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}
=\sqrt{a^2(1-\sin^2\theta)}
or
=\sqrt{a^2\cos^2\theta}
=a\cos\theta
Therefore,
\frac{x}{\sqrt{a^2-x^2}}=\frac{a\sin\theta}{a\cos\theta}
=\tan\theta
Hence, the given expression becomes
\tan^{-1}(\tan\theta)
Now,
\tan^{-1}(\tan\theta)=\theta
only when
\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Since
\theta=\sin^{-1}\left(\frac{x}{a}\right)
let
\sin \theta=\frac{x}{a}
Now, given that |x| < a, we have
\left|\frac{x}{a}\right| < 1
which implies that a real angle \theta exists such that
\sin \theta=\frac{x}{a}
Now, by definition of inverse sine function, the value of \sin^{-1}x is always taken from the principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Therefore,
\theta=\sin^{-1}\left(\frac{x}{a}\right) automatically lies in
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
because the input \frac{x}{a} lies in the valid domain (-1,1), ensuring a unique principal value exists.
Hence,
\tan^{-1}(\tan\theta)=\theta
Therefore,
\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\sin^{-1}\left(\frac{x}{a}\right)
Question 7: Ex 2.2 Inverse Trigonometric Functions
7. Simplify :
\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right),\quad a > 0,\; -\frac{a}{\sqrt3} < x < \frac{a}{\sqrt3}
Solution
In such questions, we first try to recognise whether the given expression resembles any standard trigonometric identity.
Here, the numerator and denominator resemble the expansion of
\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}
So, the natural substitution that should come to mind is
\tan\theta=\frac{x}{a}
Let
\tan\theta=\frac{x}{a}
Then,
\theta=\tan^{-1}\left(\frac{x}{a}\right)
Now,
\frac{3a^2x-x^3}{a^3-3ax^2}=\frac{3\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2}
=\tan3\theta
Therefore, the given expression becomes
\tan^{-1}(\tan3\theta)
Now,
\tan^{-1}(\tan\phi)=\phi
only when
\phi\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Since
-\frac{a}{\sqrt3} < x < \frac{a}{\sqrt3}
therefore,
-\frac1{\sqrt3} < \frac{x}{a} < \frac1{\sqrt3}
But
\tan\frac{\pi}{6}=\frac1{\sqrt3}
Hence,
-\frac{\pi}{6} < \theta < \frac{\pi}{6}
Multiplying throughout by 3, we get
-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}
Thus, 3\theta lies in the principal value branch of \tan^{-1}x.
Therefore,
\tan^{-1}(\tan3\theta)=3\theta
But
\theta=\tan^{-1}\left(\frac{x}{a}\right)
Hence,
\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)=3\tan^{-1}\left(\frac{x}{a}\right)
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move to the next question.
Question 8: Inverse Trigonometric Functions 2.2
8. Find the value of :
\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]
Solution
In such questions, we first evaluate the innermost inverse trigonometric part carefully and then simplify step-by-step using standard trigonometric identities.
Here, the inner expression is
\sin^{-1}\frac12
Now,
\sin\frac{\pi}{6}=\frac12
and \frac{\pi}{6} lies in the principal value branch of \sin^{-1}x, namely
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Therefore,
\sin^{-1}\frac12=\frac{\pi}{6}
Substituting this into the given expression, we get
\tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6}\right)\right]
=\tan^{-1}\left[2\cos\frac{\pi}{3}\right]
Now,
\cos\frac{\pi}{3}=\frac12
Therefore,
\tan^{-1}\left[2\times\frac12\right]
=\tan^{-1}(1)
Now,
\tan\frac{\pi}{4}=1
and \frac{\pi}{4} lies in the principal value branch of \tan^{-1}x, namely
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Hence,
\tan^{-1}(1)=\frac{\pi}{4}
Therefore, the value of the given expression is
\boxed{\frac{\pi}{4}}
Converting inverse trigonometric expressions into standard trigonometric equations like \sin y=x or \cos y=x often makes the correct angle and principal value easier to identify.
Question 9: Inverse Trigonometric Functions 2.2
9. Find the value of :
\tan\frac12\left[\sin^{-1}\left(\frac{2x}{1+x^2}\right)+\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right],\quad |x| < 1,\; y > 0 \text{ and } xy < 1
Solution
In such questions, the first step is to observe whether the expressions inside the inverse trigonometric functions resemble any standard trigonometric identities.
Here,
\frac{2x}{1+x^2}
resembles the identity
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}
and
\frac{1-y^2}{1+y^2}
resembles the identity
\cos2\phi=\frac{1-\tan^2\phi}{1+\tan^2\phi}
So, let
x=\tan\theta
and
y=\tan\phi
Then,
\frac{2x}{1+x^2}=\sin2\theta
and
\frac{1-y^2}{1+y^2}=\cos2\phi
Therefore, the given expression becomes
\tan\frac12\left[\sin^{-1}(\sin2\theta)+\cos^{-1}(\cos2\phi)\right]
Since |x| < 1, we have
-1 < x < 1
Now, we assumed, x=\tan\theta, therefore
-1 < \tan\theta < 1
Since tangent is increasing in its principal branch, this implies
-\frac{\pi}{4} < \theta < \frac{\pi}{4}
Hence,
-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}
So, 2\theta lies in the principal value branch of \sin^{-1}x. Therefore,
\sin^{-1}(\sin2\theta)=2\theta
Also, since y > 0, we define
\phi=\cos^{-1}y, where \cos\phi=y
Now, since y > 0, we have
0 < y \le 1
From the principal value branch of \cos^{-1}x, we know
0 \le \phi \le \pi
Since cosine is positive in the first quadrant only, and y > 0, this restricts
0 \le \phi \le \frac{\pi}{2}
Hence,
0 \le 2\phi \le \pi
which ensures that 2\phi lies in the principal value branch of \cos^{-1}x. Therefore,
\cos^{-1}(\cos2\phi)=2\phi
Therefore, the expression becomes
\tan\frac{1}{2}(2\theta+2\phi)
=\tan(\theta+\phi)
Using the identity
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
we get
\tan(\theta+\phi)=\frac{x+y}{1-xy}
Since xy < 1, the denominator is non-zero.
Hence, the value of the given expression is
\boxed{\frac{x+y}{1-xy}}
Question 10: Inverse Trigonometric Functions 2.2 Exercise
10. Find the value of :
\sin^{-1}\left(\sin\frac{2\pi}{3}\right)
Solution
In questions involving expressions like \sin^{-1}(\sin\theta), we must be very careful about the principal value branch of the inverse trigonometric function.
Many students directly cancel \sin^{-1} and \sin, but this is not always correct.
We know that the principal value branch of \sin^{-1}x is
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Now,
\frac{2\pi}{3}
does not lie in this interval.
So, we first find the sine value.
i.e. \sin\frac{2\pi}{3}=\sin\left(\pi-\frac{\pi}{3}\right)
=\sin\frac{\pi}{3}
=\frac{\sqrt3}{2}
Therefore, the given expression becomes
\sin^{-1}\left(\frac{\sqrt3}{2}\right)
Now,
\sin\frac{\pi}{3}=\frac{\sqrt3}{2}
and \frac{\pi}{3} lies in the principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Hence,
\sin^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{3}
Therefore, the value of the given expression is
\boxed{\frac{\pi}{3}}
There can be many angles having the same trigonometric value. Practising how to identify the correct angle using quadrant signs and principal value branches greatly improves conceptual understanding of inverse trigonometric functions.
Here’s the next question.
Question 11: Inverse Trigo Functions Ex. 2.2
11. Find the value of :
\tan^{-1}\left(\tan\frac{3\pi}{4}\right)
Solution
In questions involving expressions like \tan^{-1}(\tan\theta), we must always check whether the angle lies in the principal value branch of \tan^{-1}x.
You may directly cancel \tan^{-1} and \tan, but this is correct only for angles lying in the interval
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Now,
\frac{3\pi}{4}
does not lie in this interval.
So, we first evaluate the tangent value.
i.e. \tan\frac{3\pi}{4}=\tan\left(\pi-\frac{\pi}{4}\right)
=-\tan\frac{\pi}{4}
=-1
Therefore, the given expression becomes
\tan^{-1}(-1)
Now,
\tan\left(-\frac{\pi}{4}\right)=-1
and -\frac{\pi}{4} lies in the principal value branch
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Hence,
\tan^{-1}(-1)=-\frac{\pi}{4}
Therefore, the value of the given expression is
\boxed{-\frac{\pi}{4}}
Practice is the key to mastering Inverse Trigonometric Functions 2.2. Solve questions step by step and try to understand the logic behind the principal value. If you have any doubt, feel free to leave a comment.
Question 12: Inverse Trigonometric Functions 2.2
12. Find the value of :
\tan\left(\sin^{-1}\frac35+\cot^{-1}\frac32\right)
Solution
In such questions, the best approach is to evaluate each inverse trigonometric function separately and then use standard trigonometric identities.
Let
\sin^{-1}\frac35=\theta
Then,
\sin\theta=\frac35
Since \theta lies in the principal value branch of \sin^{-1}x, namely
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
therefore, \theta is acute.
Using a right triangle,
\sin\theta=\frac35
gives
\cos\theta=\frac45
Hence,
\tan\theta=\frac34
Now, let
\cot^{-1}\frac32=\phi
Then,
\cot\phi=\frac32
Therefore,
\tan\phi=\frac23
Now, the given expression becomes
\tan(\theta+\phi)
Using the identity
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
we get, \tan(\theta+\phi)=\frac{\frac34+\frac23}{1-\frac34\cdot\frac23}
=\frac{\frac{9+8}{12}}{1-\frac12}
=\frac{\frac{17}{12}}{\frac12}
or
=\frac{17}{6}
Therefore, the value of the given expression is
\boxed{\frac{17}{6}}
Now, let’s move on to the next three MCQs of Inverse Trigonometric Functions 2.2. If you want to practice or revise only the MCQs of this Chapter, you can check them here.
Question 13: Inverse Trigonometric Functions – MCQ
13. Evaluate :
\cos^{-1}\left(\cos\frac{7\pi}{6}\right)
- (A) \frac{7\pi}{6}
- (B) \frac{5\pi}{6}
- (C) \frac{\pi}{3}
- (D) \frac{\pi}{6}
Solution
In questions involving expressions like \cos^{-1}(\cos\theta), we must always check whether the angle lies in the principal value branch of \cos^{-1}x.
The principal value branch of \cos^{-1}x is
[0,\pi]
Now,
\frac{7\pi}{6}
does not lie in this interval.
So, we first evaluate the cosine value.
i.e. \cos\frac{7\pi}{6}=\cos\left(\pi+\frac{\pi}{6}\right)
=-\cos\frac{\pi}{6}
=-\frac{\sqrt3}{2}
Therefore, the given expression becomes
\cos^{-1}\left(-\frac{\sqrt3}{2}\right)
Now,
\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}
and \frac{5\pi}{6} lies in the interval
[0,\pi]
Hence,
\cos^{-1}\left(-\frac{\sqrt3}{2}\right)=\frac{5\pi}{6}
✅️ Therefore, the value of the given expression is option (B) i.e.
\boxed{\frac{5\pi}{6}}
Question 14: Inverse Trigonometric Functions 2.2 – MCQ
14. Evaluate :
\sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)\right)
- (A) \frac{1}{2}
- (B) \frac{1}{3}
- (C) \frac{1}{4}
- (D) 1
Solution
In questions involving inverse trigonometric functions, we must carefully use the principal value branches of the functions.
Now,
\sin^{-1}\left(-\frac{1}{2}\right)
means the angle lying in the principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
whose sine is -\frac{1}{2}.
Since
\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}
therefore,
\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}
Hence, \sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)\right)=\sin\left(\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right)
=\sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right)
=\sin\left(\frac{2\pi+\pi}{6}\right)
or
=\sin\left(\frac{3\pi}{6}\right)
=\sin\left(\frac{\pi}{2}\right)
i.e.
=1
✅️ Therefore, the value of the given expression is option (D) i.e.
\boxed{1}
Question 15: Inverse Trigonometric Functions 2.2 Ex – MCQ
15. Evaluate :
\tan^{-1}\sqrt3-\cot^{-1}(-\sqrt3)
- (A) \pi
- (B) -\frac{\pi}{2}
- (C) 0
- (D) 2\sqrt3
Solution
In questions involving inverse trigonometric functions, we must carefully use the principal value branches of the functions.
Now,
\tan^{-1}\sqrt3
means the angle lying in the principal value branch
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
whose tangent is \sqrt3.
Since
\tan\frac{\pi}{3}=\sqrt3
therefore,
\tan^{-1}\sqrt3=\frac{\pi}{3}
Also,
\cot^{-1}(-\sqrt3)
means the angle lying in the principal value branch of \cot^{-1}x, namely
(0,\pi)
whose cotangent is -\sqrt3.
Since
\cot\frac{5\pi}{6}=-\sqrt3
therefore,
\cot^{-1}(-\sqrt3)=\frac{5\pi}{6}
Hence, \tan^{-1}\sqrt3-\cot^{-1}(-\sqrt3)=\frac{\pi}{3}-\frac{5\pi}{6}
=\frac{2\pi-5\pi}{6}
=-\frac{3\pi}{6}
i.e.
=-\frac{\pi}{2}
✅️ Therefore, the value of the given expression is option (B) i.e.
\boxed{-\frac{\pi}{2}}
Common Mistakes to Avoid
- Direct substitution without converting inverse to angle: Students often try to simplify expressions like \sin\left(\frac{\pi}{3}-\sin^{-1}x\right) directly without first converting \sin^{-1}x into a principal angle. This leads to confusion in simplification.
- Incorrect handling of nested inverse expressions: Expressions like \sin^{-1}(\sin x) or \cos^{-1}(\cos x) are often simplified to x without checking whether x lies in the principal value range.
- Not reducing angles properly before applying trig functions: Students sometimes skip simplifying angles like \frac{\pi}{3}+\frac{\pi}{6} before applying sine or cosine, leading to avoidable errors.
- Sign errors in inverse trigonometric values: A common mistake is missing negative principal values such as \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}.
- Incorrect use of compound angle formulas: While simplifying expressions like \sin(a-b), students often apply formulas incorrectly or forget sign changes.
- Assuming inverse functions cancel freely: Expressions like \sin(\sin^{-1}x) are assumed to always equal x, without checking domain restrictions.
- Mixing up principal value substitution steps: Students often jump directly to final answers without explicitly identifying the principal value of inverse trig functions.
- Errors in angle addition/subtraction inside sine and cosine: Misapplication of identities like \sin(A\pm B) is common when inverse trigonometric expressions are involved.
- Not checking final angle validity: Even after simplification, students sometimes forget to verify if intermediate inverse steps respected their principal value ranges.
- Over-reliance on memorized results: Students try to recall direct answers instead of actually converting inverse expressions step-by-step, which leads to errors in non-standard forms.
Continue Learning
After completing Exercise 2.2 of Inverse Trigonometric Functions, you should now be comfortable with simplifying expressions involving inverse trigonometric functions using principal value concepts, standard identities, and trigonometric transformations.
To strengthen your understanding further, make sure that you revise:
- Evaluation of expressions involving combinations of trigonometric and inverse trigonometric functions
- Careful use of principal value branches while simplifying expressions
- Conversion of inverse trigonometric expressions into standard angles before simplification
- Application of identities like \sin(\sin^{-1}x), \cos(\cos^{-1}x), \tan(\tan^{-1}x) with domain restrictions
- Use of compound angle formulas such as \sin(A \pm B), \cos(A \pm B) in inverse trigonometric expressions
- Handling expressions where inverse trigonometric functions appear inside other trigonometric functions
- Identifying and simplifying nested inverse trigonometric expressions correctly
- Ensuring final answers lie within correct principal value ranges
Explore More
Try creating your own examples involving expressions with inverse trigonometric functions and practice simplifying them step-by-step using principal value branches. Focus on converting inverse trigonometric expressions into standard angles and then applying trigonometric identities for simplification.
You should also verify your final answers by checking whether the resulting angle lies within the correct principal value range. This will help avoid common errors and strengthen conceptual clarity in evaluating such expressions.
All the best and keep learning 👍
