CBSE Class 12 Mathematics Inverse Trigonometric Functions Worksheet Set 02

Question. If sin-1 x – cos-1 x = 𝜋/6, then x =
(a) 1/2
(b) √3/2
(c) −1/2
(d) −√3/2
Answer : B

Question. If tan-1 (cot θ) = 2θ, then θ is equal to
(a) 𝜋/3
(b) 𝜋/4
(c) 𝜋/6
(d) None of these
Answer : C

Question. cot( 𝜋/4 – 2 cot-1 3) =
(a) 7
(b) 6
(c) 5
(d) None of these
Answer : A

Question. The principal value of tan-1(tan 3π/5) is
(a) 2π/5
(b) -2π/5
(c) 3π/5
(d) -3π/5
Answer : B

Question. sin[π/3 – sin^-1(- ½)] is equal to:
(a) 1/2
(b) 1/3
(c) -1
(d) 1
Answer : D

Question. The domain of sin^–1(2x) is
(a) [0, 1]
(b) [– 1, 1]
(c) [-1/2, 1/2]
(d) [–2, 2]
Answer : C

Question. If sin^–1 x + sin^–1 y = π/2, then value of cos^–1 x + cos^–1 y is
(a) π/2
(b) π
(c) 0
(d) 2π/3
Answer : A

Question. The domain of y = cos^–1 (x² – 4) is
(a) [3, 5]
(b) [0, π]
(c) [-√5, -√3] ∩ [-√5, √3]
(d) [-√5, -√3] ∪ [√3, √5]
Answer : D

Question. The value of the expression sin [cot^–1 (cos (tan^–1 1))] is
(a) 0
(b) 1
(c) 1/√3
(d) √2/√3
Answer : D

Question. Which of the following is the principal value branch of cos^–1 x?
(a) [–π/2, π/2]
(b) (0, π)
(c) [0, π]
(d) (0, π) – {π/2}
Answer : C

CASE BASED STUDY QUESTIONS

1. Read the following text and answer the following questions on the basis of the same:
In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan-1(½) and tan-1(⅓) respectively.

Question. The value of sin A is _______.
A. 1/2
B. 1/3
C. 1/√5
D. 2/√5
Answer : C

Question. The third angle, ∠C = _______.
A. π/4
B. π/2
C. π/3
D. 3π/4
Answer : D

Question. cos (A + B +C)
A. 1
B. 0
C. – 1
D. 1/2
Answer : C

Question. If A = sin-1 x , then the value of x is
A. 1/√10
B. 3/√10
C. 1/√5
D. 2/√5
Answer : C

Question. If A = cos-1 x , then the value of x is
A. 1/√10
B. 3/√10
C. 1/√5
D. 2/√5
Answer : B

2. The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. ‘A’ is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby, Ram Robert and Rahim suggested to the film to place the hoarding board at three different locations namely C, D and E. ‘C’ is at the height of 10 metres from the ground level. For the viewer ‘A’, the angle of elevation of ‘D’ is double the angle of elevation of ‘C’. The angle of elevation of ‘E’ is triple the angle of elevation of ‘C’ for the same viewer.
Look at the figure given and based on the above information answer the following:

Question. Measure of ∠CAB =
(a) tan–1(2)
(b) tan–1(1/2)
(c) tan–1(1)
(d) tan–1(3)
Answer : B

Question. Measure of ∠DAB =
(a) tan–1(3/4)
(b) tan–1(3)
(c) tan–1(4/3)
(d) tan–1(4)
Answer : C

Question. Measure of ∠EAB
(a) tan–1(11)
(b) tan–1(3)
(c) tan–1(2/11)
(d) tan–1(11/2)
Answer : D

Question. A’ is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠CAB and ∠CA’B is
(a) tan–1(1/12)
(b) tan–1(1/8)
(c) tan–1(2/5)
(d) tan–1(11/21)
Answer : A

Please click the link below to download full pdf file for CBSE Class 12 Mathematics Worksheet - Inverse Trigonometric Functions.

Page 15

Topic 2

Inverse Trigonometric Functions

Schematic Diagram

Some Important Results/Concepts

* Domain & Range of the Inverse Trigonometric Function:

* Properties of Inverse Trigonometric Functions:

1. Self-Inverse Properties:

• (i) \( \sin^{-1}(\sin x) = x \) and \( \sin(\sin^{-1} x) = x \)
• (ii) \( \cos^{-1}(\cos x) = x \) and \( \cos(\cos^{-1} x) = x \)
• (iii) \( \tan^{-1}(\tan x) = x \) and \( \tan(\tan^{-1} x) = x \)
• (iv) \( \cot^{-1}(\cot x) = x \) and \( \cot(\cot^{-1} x) = x \)
• (v) \( \sec^{-1}(\sec x) = x \) and \( \sec(\sec^{-1} x) = x \)
• (vi) \( \csc^{-1}(\csc x) = x \) and \( \csc(\csc^{-1} x) = x \)

2. Reciprocal Relations:

• (i) \( \sin^{-1} x = \csc^{-1} \frac{1}{x} \) & \( \csc^{-1} x = \sin^{-1} \frac{1}{x} \)
• (ii) \( \cos^{-1} x = \sec^{-1} \frac{1}{x} \) & \( \sec^{-1} x = \cos^{-1} \frac{1}{x} \)
• (iii) \( \tan^{-1} x = \cot^{-1} \frac{1}{x} \) & \( \cot^{-1} x = \tan^{-1} \frac{1}{x} \)

3. Negative Argument Properties:

• (i) \( \sin^{-1}(-x) = -\sin^{-1} x \)
• (ii) \( \tan^{-1}(-x) = -\tan^{-1} x \)
• (iii) \( \csc^{-1}(-x) = -\csc^{-1} x \)
• (iv) \( \cos^{-1}(-x) = \pi - \cos^{-1} x \)
• (v) \( \sec^{-1}(-x) = \pi - \sec^{-1} x \)
• (vi) \( \cot^{-1}(-x) = \pi - \cot^{-1} x \)

4. Complementary Identities:

• (i) \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \)
• (ii) \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \)
• (iii) \( \csc^{-1} x + \sec^{-1} x = \frac{\pi}{2} \)

Page 16

5. Double-Angle Transformations:

\[ 2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]

6. Sum and Difference Identities:

• \( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) if \( xy < 1 \)
• \( \tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) if \( xy > 1 \)
• \( \tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \) if \( xy > -1 \)

Assignments

(i). Principal value branch Table

Level I

Write the principal value of the following :

Question 1. \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
Answer: Let \( y = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \).
This equation can be rewritten as:
\( \cos y = \frac{\sqrt{3}}{2} \)
Since the principal value branch of the inverse cosine function is restricted to \( [0, \pi] \), we choose the angle in this interval:
\( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
Thus, the principal value is \( \frac{\pi}{6} \).
In simple words: Find the angle between 0 and 180 degrees where the cosine value is \(\sqrt{3}/2\). That angle is 30 degrees, which is written as \(\pi/6\).

Exam Tip: Always make sure your final angle lies within the standard principal branch of the given inverse function.

Question 2. \( \sin^{-1}\left(-\frac{1}{2}\right) \)
Answer: Let \( y = \sin^{-1}\left(-\frac{1}{2}\right) \).
This equation can be rewritten as:
\( \sin y = -\frac{1}{2} \)
The principal value branch of the inverse sine function is restricted to \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Within this interval, we identify:
\( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \)
Consequently, the principal value is \( -\frac{\pi}{6} \).
In simple words: We need to find the angle between -90 and 90 degrees that has a sine value of -1/2. That angle is -30 degrees, or -\(\pi/6\).

Exam Tip: Do not use positive angles like \( 11\pi/6 \) for inverse sine calculations, as they lie outside the standard range of \( [-\pi/2, \pi/2] \).

Question 3. \( \tan^{-1}(-\sqrt{3}) \)
Answer: Let \( y = \tan^{-1}(-\sqrt{3}) \).
This expression simplifies to:
\( \tan y = -\sqrt{3} \)
The principal value branch of the inverse tangent function is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). We choose the angle in this open interval:
\( \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \)
This gives us the principal value of \( -\frac{\pi}{3} \).
In simple words: Look for the angle between -90 and 90 degrees where tangent is equal to -\(\sqrt{3}\). That angle is -60 degrees, which is written as -\(\pi/3\).

Exam Tip: Remember that inverse tangent has an open interval range, so the boundary values of \( -\pi/2 \) and \( \pi/2 \) are excluded.

Question 4. \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
Answer: Let \( y = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \).
This can be written as:
\( \cos y = -\frac{1}{\sqrt{2}} \)
The principal value branch of inverse cosine is \( [0, \pi] \). We can use the identity \( \cos^{-1}(-x) = \pi - \cos^{-1} x \) to find the value:
\( y = \pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \)

\( \implies y = \pi - \frac{\pi}{4} \)

\( \implies y = \frac{3\pi}{4} \).
Thus, the principal value is \( \frac{3\pi}{4} \).
In simple words: For negative cosine arguments, find the reference angle first (45 degrees) and subtract it from 180 degrees. This gives 135 degrees, which is \(3\pi/4\).

Exam Tip: Negative inputs in inverse cosine always result in an obtuse angle in the second quadrant (between \( \pi/2 \) and \( \pi \)).

Level II

Write the principal value of the following :

Question 1. \( \cos^{-1}\left(\cos \frac{2\pi}{3}\right) + \sin^{-1}\left(\sin \frac{2\pi}{3}\right) \) 
Answer: We calculate both components individually:
For the first term, \( \cos^{-1}\left(\cos \frac{2\pi}{3}\right) \):
The angle \( \frac{2\pi}{3} \) lies within the principal value branch of \( \cos^{-1} \), which is \( [0, \pi] \). Thus:
\( \cos^{-1}\left(\cos \frac{2\pi}{3}\right) = \frac{2\pi}{3} \).

For the second term, \( \sin^{-1}\left(\sin \frac{2\pi}{3}\right) \):
The angle \( \frac{2\pi}{3} \) does not fall in the principal value branch of \( \sin^{-1} \), which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). We rewrite the term using trigonometric identities:
\( \sin \frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \sin \frac{\pi}{3} \).
Thus:
\( \sin^{-1}\left(\sin \frac{2\pi}{3}\right) = \sin^{-1}\left(\sin \frac{\pi}{3}\right) = \frac{\pi}{3} \).

Now, we sum these two results:
\( \text{Sum} = \frac{2\pi}{3} + \frac{\pi}{3} = \frac{3\pi}{3} = \pi \).
The principal value of the entire expression is \( \pi \).
In simple words: The first part is already within the correct range, so it stays as 120 degrees. The second part must be converted to 60 degrees to fit the sine range. Adding 120 and 60 degrees gives 180 degrees, which is \(\pi\).

Exam Tip: Be careful not to just cancel \( \sin^{-1} \) and \( \sin \) when the angle is outside the \( [-\pi/2, \pi/2] \) interval. Convert it first.

Question 2. \( \sin^{-1}\left(\sin \frac{4\pi}{5}\right) \)
Answer: We evaluate the expression:
The angle \( \frac{4\pi}{5} \) does not lie in the principal value branch of \( \sin^{-1} \), which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). We must find an equivalent angle in the correct range.
Using the identity \( \sin \theta = \sin(\pi - \theta) \):
\( \sin \frac{4\pi}{5} = \sin\left(\pi - \frac{4\pi}{5}\right) = \sin \frac{\pi}{5} \).
Since \( \frac{\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), we can now simplify:
\( \sin^{-1}\left(\sin \frac{4\pi}{5}\right) = \sin^{-1}\left(\sin \frac{\pi}{5}\right) = \frac{\pi}{5} \).
Thus, the principal value is \( \frac{\pi}{5} \).
In simple words: Since 144 degrees is outside the allowable range for sine, we use the identity to convert it to 36 degrees, which is \(\pi/5\).

Exam Tip: For any angle \( \theta \) in the second quadrant, write \( \sin \theta \) as \( \sin(\pi - \theta) \) to bring it into the principal branch.

Question 3. \( \cos^{-1}\left(\cos \frac{7\pi}{6}\right) \)
Answer: We evaluate the expression:
The angle \( \frac{7\pi}{6} \) does not fall in the principal value branch of \( \cos^{-1} \), which is \( [0, \pi] \). We must find an equivalent angle in this range.
Using the identity \( \cos \theta = \cos(2\pi - \theta) \):
\( \cos \frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos \frac{5\pi}{6} \).
Since \( \frac{5\pi}{6} \in [0, \pi] \), we can now simplify:
\( \cos^{-1}\left(\cos \frac{7\pi}{6}\right) = \cos^{-1}\left(\cos \frac{5\pi}{6}\right) = \frac{5\pi}{6} \).
Thus, the principal value is \( \frac{5\pi}{6} \).
In simple words: 210 degrees is too large for the cosine range. We find its equivalent in the first two quadrants, which is 150 degrees, or \(5\pi/6\).

Exam Tip: For angles in the third quadrant, use \( \cos \theta = \cos(2\pi - \theta) \) to shift the angle to the second quadrant.

(ii). Properties of Inverse Trigonometric Functions

Level I

Question 1. Evaluate \( \cot\left[\tan^{-1} a + \cot^{-1} a\right] \)
Answer: We use the complementary angle identity for inverse trigonometric functions:
\( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \) for all real values of \( x \).
Substituting this identity into our expression gives:
\( \cot\left[\tan^{-1} a + \cot^{-1} a\right] = \cot\left(\frac{\pi}{2}\right) \).
We know that \( \cot\left(\frac{\pi}{2}\right) = 0 \).
Thus, the evaluated result is 0.
In simple words: The sum of inverse tangent and inverse cotangent of the same number is always 90 degrees. The cotangent of 90 degrees is 0.

Exam Tip: Memorizing the complementary angle sums (\( \pi/2 \)) for inverse trig pairs makes solving these types of problems extremely fast.

Question 2. Prove \( 3\sin^{-1} x = \sin^{-1}(3x - 4x^3) \)
Answer: Let \( x = \sin \theta \implies \theta = \sin^{-1} x \).
We substitute this value into the Right Hand Side (RHS) of the equation:
\( \text{RHS} = \sin^{-1}(3x - 4x^3) \)

\( \implies \text{RHS} = \sin^{-1}(3\sin \theta - 4\sin^3 \theta) \)
Using the triple-angle identity \( \sin 3\theta = 3\sin \theta - 4\sin^3 \theta \):
\( \text{RHS} = \sin^{-1}(\sin 3\theta) \)

\( \implies \text{RHS} = 3\theta \)
Substituting back \( \theta = \sin^{-1} x \):
\( \text{RHS} = 3\sin^{-1} x = \text{LHS} \).
Hence, proved.
In simple words: Substitute x with sine of theta. The formula inside the bracket is the formula for the sine of three-theta. The inverse sine cancels this out, leaving us with three-theta, which equals the left side.

Exam Tip: Whenever you see the expression \( 3x - 4x^3 \), it is a strong hint to substitute \( x = \sin \theta \).

Question 3. Find x if \( \sec^{-1}(\sqrt{2}) + \csc^{-1} x = \frac{\pi}{2} \)
Answer: We use the standard complementary identity:
\( \sec^{-1} y + \csc^{-1} y = \frac{\pi}{2} \) for all \( |y| \ge 1 \).
Comparing this identity with the given equation:
\( \sec^{-1}(\sqrt{2}) + \csc^{-1} x = \frac{\pi}{2} \)
By direct comparison, the arguments of the two functions must be equal for the sum to be \( \frac{\pi}{2} \).
Therefore, we get:
\( x = \sqrt{2} \).
In simple words: Since inverse secant and inverse cosecant of the same number must add up to 90 degrees, x must be equal to the number inside the first function, which is \(\sqrt{2}\).

Exam Tip: Recognizing the direct comparison method can save you from doing unnecessary calculations on exams.

Level II

Question 1. Write the following in simplest form : \( \tan^{-1}\left(\frac{\sqrt{1+x^2} - 1}{x}\right) , x \neq 0 \)
Answer: Let \( x = \tan \theta \implies \theta = \tan^{-1} x \).
We substitute this into the expression:
\[ E = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta} - 1}{\tan \theta}\right) \]
Since \( 1 + \tan^2 \theta = \sec^2 \theta \), we get:
\[ E = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) \]
Converting the terms to sine and cosine:
\[ E = \tan^{-1}\left(\frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}}\right) \]

\[ \implies E = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) \]
Using half-angle trigonometric formulas:
\( 1 - \cos \theta = 2\sin^2 \frac{\theta}{2} \) and \( \sin \theta = 2\sin \frac{\theta}{2}\cos \frac{\theta}{2} \).
Substituting these in:
\[ E = \tan^{-1}\left(\frac{2\sin^2 \frac{\theta}{2}}{2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}\right) \]

\[ \implies E = \tan^{-1}\left(\tan \frac{\theta}{2}\right) \]

\[ \implies E = \frac{\theta}{2} \].
Replacing \( \theta \) back with \( \tan^{-1} x \):
\( E = \frac{1}{2}\tan^{-1} x \).
This is the simplest form of the expression.
In simple words: Substitute x as tangent of theta. Convert the algebra into sine and cosine terms, use half-angle formulas to simplify the fraction to tangent of half-theta, which reduces to half of inverse tangent of x.

Exam Tip: Substituting trigonometric functions like tangent is highly effective for simplifying expressions containing \( \sqrt{1+x^2} \).

Page 17

Question 2. Prove that \( \sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36} \)
Answer: Let \( \sin^{-1} \frac{8}{17} = A \implies \sin A = \frac{8}{17} \).
We find \( \cos A \):
\( \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \sqrt{\frac{289-64}{289}} = \frac{15}{17} \).
Thus, \( \tan A = \frac{\sin A}{\cos A} = \frac{8}{15} \).

Let \( \sin^{-1} \frac{3}{5} = B \implies \sin B = \frac{3}{5} \).
We find \( \cos B \):
\( \cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \).
Thus, \( \tan B = \frac{3}{4} \).

Using the tangent addition formula:
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

\( \implies \tan(A+B) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}} \)

\( \implies \tan(A+B) = \frac{\frac{32 + 45}{60}}{1 - \frac{24}{60}} \)

\( \implies \tan(A+B) = \frac{\frac{77}{60}}{\frac{36}{60}} = \frac{77}{36} \).
Therefore:
\( A + B = \tan^{-1}\left(\frac{77}{36}\right) \)

\( \implies \sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36} \).
Hence, proved.
In simple words: Convert both inverse sine terms into inverse tangent terms by drawing reference triangles. Then use the tangent sum formula to calculate the combined angle.

Exam Tip: Converting different inverse trigonometric functions to inverse tangent is a very reliable strategy for proving identities.

Question 3. Prove that \( \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{8} = \frac{\pi}{4} \)
Answer: We group the four terms into two pairs:
\( \text{LHS} = \left[\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right)\right] + \left[\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right)\right] \).
Using the sum formula \( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \):
For the first group:
\( \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{1/3 + 1/5}{1 - 1/15}\right) = \tan^{-1}\left(\frac{8/15}{14/15}\right) = \tan^{-1}\left(\frac{4}{7}\right) \).

For the second group:
\( \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{1/7 + 1/8}{1 - 1/56}\right) = \tan^{-1}\left(\frac{15/56}{55/56}\right) = \tan^{-1}\left(\frac{3}{11}\right) \).

Now, we sum the two simplified terms:
\( \text{LHS} = \tan^{-1}\left(\frac{4}{7}\right) + \tan^{-1}\left(\frac{3}{11}\right) \)

\( \implies \text{LHS} = \tan^{-1}\left(\frac{4/7 + 3/11}{1 - 12/77}\right) \)

\( \implies \text{LHS} = \tan^{-1}\left(\frac{\frac{44 + 21}{77}}{\frac{77 - 12}{77}}\right) \)

\( \implies \text{LHS} = \tan^{-1}\left(\frac{65}{65}\right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{RHS} \).
Hence, proved.
In simple words: Pair up the four terms and combine them two at a time using the formula. Combine the two resulting fractions once more to get 1, which has an inverse tangent of 45 degrees (\(\pi/4\)).

Exam Tip: Grouping terms systematically keeps the arithmetic clean and reduces the chances of fraction mistakes.

Question 4. Prove that \( 2\tan^{-1} \left(\frac{1}{2}\right) + \tan^{-1} \left(\frac{1}{7}\right) = \tan^{-1} \left(\frac{31}{17}\right) \) [CBSE 2011]
Answer: First, we simplify the double-angle term using the identity \( 2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \):
\( 2\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{2(1/2)}{1 - (1/2)^2}\right) = \tan^{-1}\left(\frac{1}{1 - 1/4}\right) = \tan^{-1}\left(\frac{4}{3}\right) \).

Now we substitute this back into the LHS of the equation:
\( \text{LHS} = \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
Using the tangent sum formula:
\( \text{LHS} = \tan^{-1}\left(\frac{4/3 + 1/7}{1 - \frac{4}{3} \cdot \frac{1}{7}}\right) \)

\( \implies \text{LHS} = \tan^{-1}\left(\frac{\frac{28 + 3}{21}}{\frac{21 - 4}{21}}\right) \)

\( \implies \text{LHS} = \tan^{-1}\left(\frac{31}{17}\right) = \text{RHS} \).
Hence, proved.
In simple words: First double the first angle using the identity formula to turn it into an inverse tangent of 4/3. Then add the second angle to it to get the final fraction of 31/17.

Exam Tip: Always resolve the coefficient (like 2 in \( 2\tan^{-1} x \)) before attempting to combine terms using the standard sum formula.

Question 5. Prove that \( \sin^{-1} \left(\frac{8}{17}\right) + \sin^{-1} \left(\frac{3}{5}\right) = \cos^{-1} \left(\frac{36}{85}\right) \) [CBSE 2012]
Answer: Let \( \sin^{-1} \frac{8}{17} = A \implies \sin A = \frac{8}{17} \), and \( \cos A = \frac{15}{17} \).
Let \( \sin^{-1} \frac{3}{5} = B \implies \sin B = \frac{3}{5} \), and \( \cos B = \frac{4}{5} \).
We use the cosine addition identity:
\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
Substituting our known trigonometric values in:
\( \cos(A+B) = \left(\frac{15}{17}\right)\left(\frac{4}{5}\right) - \left(\frac{8}{17}\right)\left(\frac{3}{5}\right) \)

\( \implies \cos(A+B) = \frac{60}{85} - \frac{24}{85} \)

\( \implies \cos(A+B) = \frac{36}{85} \).
Taking the inverse cosine on both sides:
\( A + B = \cos^{-1}\left(\frac{36}{85}\right) \)
Substituting back the initial terms:
\( \sin^{-1} \left(\frac{8}{17}\right) + \sin^{-1} \left(\frac{3}{5}\right) = \cos^{-1} \left(\frac{36}{85}\right) \).
Hence, proved.
In simple words: Define the two terms as angles A and B, find their cosine values, and plug them into the cosine of (A + B) formula to verify the exact match.

Exam Tip: Using the cosine addition formula directly is much faster than converting everything to inverse cosine first.

Level III

Question 1. Prove that \( \cot^{-1} \left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2} , x \in \left(0, \frac{\pi}{4}\right) \)
Answer: We simplify the terms inside the square root first.
We know that:
\( 1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2} = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2 \).
Thus:
\( \sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2} \) (as both terms are positive since \( x \in (0, \pi/4) \)).

Similarly:
\( \sqrt{1-\sin x} = \cos \frac{x}{2} - \sin \frac{x}{2} \) (since \( \cos \frac{x}{2} > \sin \frac{x}{2} \) in the given domain).

Substituting these expressions into our main formula:
Numerator: \( \sqrt{1+\sin x} + \sqrt{1-\sin x} = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right) + \left(\cos \frac{x}{2} - \sin \frac{x}{2}\right) = 2\cos \frac{x}{2} \).
Denominator: \( \sqrt{1+\sin x} - \sqrt{1-\sin x} = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right) - \left(\cos \frac{x}{2} - \sin \frac{x}{2}\right) = 2\sin \frac{x}{2} \).

Now, we write:
\( \text{LHS} = \cot^{-1}\left(\frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}}\right) \)

\( \implies \text{LHS} = \cot^{-1}\left(\cot \frac{x}{2}\right) \)

\( \implies \text{LHS} = \frac{x}{2} = \text{RHS} \).
Hence, proved.
In simple words: Rewrite the term 1 as \(\cos^2(x/2) + \sin^2(x/2)\) and \(\sin(x)\) as the double-angle sine term. This turns the square roots into perfect squares that cancel out, simplifying the fraction to cotangent of x/2.

Exam Tip: Pay attention to the domain of x. In this case, \( \cos(x/2) > \sin(x/2) \), so we must write \( \sqrt{1-\sin x} \) as \( \cos(x/2) - \sin(x/2) \).

Question 2. Prove that \( \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x \) [CBSE 2011]
Answer: Let \( x = \cos 2\theta \implies 2\theta = \cos^{-1} x \implies \theta = \frac{1}{2}\cos^{-1} x \).
We substitute this into the LHS:
\( \sqrt{1+x} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2 \theta} = \sqrt{2}\cos \theta \).
\( \sqrt{1-x} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2 \theta} = \sqrt{2}\sin \theta \).
Substituting these values into the expression:
\[ \text{LHS} = \tan^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right) \]

\[ \implies \text{LHS} = \tan^{-1}\left(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\right) \]
Divide the numerator and denominator by \( \cos \theta \):
\[ \text{LHS} = \tan^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right) \]
Using the identity \( \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1-\tan \theta}{1+\tan \theta} \):
\[ \text{LHS} = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \]

\[ \implies \text{LHS} = \frac{\pi}{4} - \theta \]
Substituting back \( \theta = \frac{1}{2}\cos^{-1} x \):
\( \text{LHS} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x = \text{RHS} \).
Hence, proved.
In simple words: Substituting x with cos(2-theta) converts the square root expressions into basic sine and cosine functions. Simplifying this matches the target formula of 45 degrees minus theta.

Exam Tip: Substituting \( x = \cos 2\theta \) is the most efficient way to simplify expressions containing the term \( \sqrt{1 \pm x} \).

Question 3. Solve \( \tan^{-1} 2x + \tan^{-1} 3x = \pi / 4 \)
Answer: We apply the tangent sum formula to the left side:
\( \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) = \frac{\pi}{4} \)

\( \implies \frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right) \)

\( \implies \frac{5x}{1 - 6x^2} = 1 \)

\( \implies 5x = 1 - 6x^2 \)

\( \implies 6x^2 + 5x - 1 = 0 \).
Solving the quadratic equation by factoring:
\( 6x^2 + 6x - x - 1 = 0 \)

\( \implies 6x(x+1) - 1(x+1) = 0 \)

\( \implies (6x-1)(x+1) = 0 \)
This gives two possible roots: \( x = \frac{1}{6} \) or \( x = -1 \).

We must check if the solutions are valid:
If \( x = -1 \), both \( \tan^{-1}(2x) \) and \( \tan^{-1}(3x) \) are negative angles, so their sum cannot be positive \( \pi/4 \). Thus, \( x = -1 \) is an extraneous solution.
For \( x = \frac{1}{6} \), the sum of positive angles is positive and satisfies the equation.
Thus, the only valid solution is \( x = \frac{1}{6} \).
In simple words: Combine the terms into a single fraction and solve the quadratic equation. Make sure to throw away the negative solution because it doesn't satisfy the original equation.

Exam Tip: Always plug your solutions back into inverse trig equations to check for extraneous negative roots.

Question 4. Solve \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{8}{31}\right) \)
Answer: Applying the tangent sum formula:
\( \tan^{-1}\left(\frac{(x+1) + (x-1)}{1 - (x+1)(x-1)}\right) = \tan^{-1}\left(\frac{8}{31}\right) \)

\( \implies \frac{2x}{1 - (x^2 - 1)} = \frac{8}{31} \)

\( \implies \frac{2x}{2 - x^2} = \frac{8}{31} \)

\( \implies \frac{x}{2 - x^2} = \frac{4}{31} \).
Cross-multiplying gives:
\( 31x = 4(2 - x^2) \)

\( \implies 31x = 8 - 4x^2 \)

\( \implies 4x^2 + 31x - 8 = 0 \).
Factoring the quadratic equation:
\( 4x^2 + 32x - x - 8 = 0 \)

\( \implies 4x(x+8) - 1(x+8) = 0 \)

\( \implies (4x-1)(x+8) = 0 \).
This yields \( x = \frac{1}{4} \) or \( x = -8 \).

Checking the solutions:
For \( x = -8 \), the left side contains negative arguments which sum to a negative value, while the right side is positive. Thus, \( x = -8 \) is invalid.
Therefore, the only valid solution is \( x = \frac{1}{4} \).
In simple words: Apply the formula to combine the tangents, cross-multiply the fractions, and factor the quadratic equation. Discard the negative solution of -8.

Exam Tip: Double check that your final solutions do not make the denominator \( 1-xy \) negative, unless \( \pi \) is adjusted for accordingly.

Question 5. Solve \( \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4} \)
Answer: We use the tangent sum formula:
\( \tan^{-1}\left( \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)} \right) = \frac{\pi}{4} \)
Taking the tangent on both sides:
\( \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan\left(\frac{\pi}{4}\right) \)

\( \implies \frac{(x^2 + x - 2) + (x^2 - x - 2)}{(x^2 - 4) - (x^2 - 1)} = 1 \)

\( \implies \frac{2x^2 - 4}{-3} = 1 \)

\( \implies 2x^2 - 4 = -3 \)

\( \implies 2x^2 = 1 \)

\( \implies x^2 = \frac{1}{2} \)

\( \implies x = \pm \frac{1}{\sqrt{2}} \).
Both values of \( x \) satisfy the initial equation.
Thus, the solutions are \( x = \pm \frac{1}{\sqrt{2}} \).
In simple words: Combine the fractions, simplify the resulting algebra, and solve for x. This gives two valid solutions, positive and negative 1 divided by the square root of 2.

Exam Tip: Be careful when simplifying the denominator: \( (x^2-4)-(x^2-1) \) simplifies to \( -3 \), not \( -5 \).

Question 6. Prove that \( \tan^{-1} \left(\frac{\cos x}{1+\sin x}\right) = \frac{\pi}{4} - \frac{x}{2} , x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) [CBSE 2012]
Answer: We rewrite the trigonometric terms using half-angle identities:
We know that:
\( \cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \).
And since \( 1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \) and \( \sin x = 2\sin \frac{x}{2}\cos \frac{x}{2} \):
\( 1 + \sin x = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2 \).
Substituting these into the expression:
\[ \text{LHS} = \tan^{-1}\left(\frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2}\right) \]
Factoring the numerator as a difference of squares:
\[ \text{LHS} = \tan^{-1}\left(\frac{\left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2}\right) \]

\[ \implies \text{LHS} = \tan^{-1}\left(\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}\right) \]
Dividing by \( \cos \frac{x}{2} \):
\[ \text{LHS} = \tan^{-1}\left(\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}\right) \]

\[ \implies \text{LHS} = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) \]

\[ \implies \text{LHS} = \frac{\pi}{4} - \frac{x}{2} = \text{RHS} \].
Hence, proved.
In simple words: Rewrite cos(x) and sin(x) in terms of half-angles, factor the numerator to cancel common terms, and use the tangent subtraction formula to simplify the expression.

Exam Tip: Substituting \( \cos x = \sin(\pi/2 - x) \) and \( \sin x = \cos(\pi/2 - x) \) is another valid method to solve this identity.

Questions for self evaluation

Question 1. Prove that \( \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5} = \tan^{-1} \frac{63}{16} \)
Answer: Let \( \sin^{-1} \frac{5}{13} = A \implies \sin A = \frac{5}{13} \).
We calculate \( \cos A = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13} \).
Thus, \( \tan A = \frac{5}{12} \).

Let \( \cos^{-1} \frac{3}{5} = B \implies \cos B = \frac{3}{5} \).
We calculate \( \sin B = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \).
Thus, \( \tan B = \frac{4}{3} \).

Using the tangent addition formula:
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

\( \implies \tan(A+B) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \cdot \frac{4}{3}} \)

\( \implies \tan(A+B) = \frac{\frac{5 + 16}{12}}{1 - \frac{20}{36}} \)

\( \implies \tan(A+B) = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{\frac{7}{4}}{\frac{4}{9}} = \frac{63}{16} \).
Therefore:
\( A + B = \tan^{-1}\left(\frac{63}{16}\right) \).
This proves the identity.
In simple words: Convert both inverse functions into tangent values using basic trigonometry. Plug those fractions into the tangent addition formula to confirm the target value.

Exam Tip: Always write down your coordinate conversions explicitly so you do not make simple calculation mistakes.

Question 2. Prove that \( \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x \) , \( x \in \left[-\frac{1}{\sqrt{2}}, 1\right] \)
Answer: Let \( x = \cos 2\theta \implies \theta = \frac{1}{2}\cos^{-1} x \).
We rewrite the terms using trigonometric identities:
\( \sqrt{1+\cos 2\theta} = \sqrt{2}\cos \theta \).
\( \sqrt{1-\cos 2\theta} = \sqrt{2}\sin \theta \).
Substituting these values into the left side of the equation:
\[ \text{LHS} = \tan^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right) \]

\[ \implies \text{LHS} = \tan^{-1}\left(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\right) \]
Divide the numerator and denominator by \( \cos \theta \):
\[ \text{LHS} = \tan^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right) \]

\[ \implies \text{LHS} = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) \]

\[ \implies \text{LHS} = \frac{\pi}{4} - \theta \]
Replacing \( \theta \) back with \( \frac{1}{2}\cos^{-1} x \):
\( \text{LHS} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x = \text{RHS} \).
Hence, proved.
In simple words: Substituting x with the cosine function lets you use identities to clear the square roots. Dividing by cosine turns the fraction into tangent of (45 degrees - theta), which matches the target formula.

Exam Tip: Make sure you state the valid domain interval for the substitution at each stage of your proof.

Question 3. Prove that \( \sin^{-1} \frac{12}{13} + \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{63}{16} = \pi \)
Answer: Let us convert all terms to inverse tangents:
Let \( \sin^{-1} \frac{12}{13} = A \implies \sin A = \frac{12}{13} \implies \tan A = \frac{12}{5} \).
Let \( \cos^{-1} \frac{4}{5} = B \implies \cos B = \frac{4}{5} \implies \tan B = \frac{3}{4} \).
Let \( \tan^{-1} \frac{63}{16} = C \implies \tan C = \frac{63}{16} \).
We combine \( A \) and \( B \) using the tangent sum formula:
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{12/5 + 3/4}{1 - \frac{12}{5} \cdot \frac{3}{4}} = \frac{63/20}{-16/20} = -\frac{63}{16} \).
Since \( \tan(A+B) = -\tan C \) and both \( A \) and \( B \) are positive acute angles (so \( A+B > \frac{\pi}{2} \)):
\( A + B = \pi - C \)

\( \implies A + B + C = \pi \).
Thus, \( \sin^{-1} \frac{12}{13} + \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{63}{16} = \pi \).
Hence, proved.
In simple words: Convert the first two angles to tangent values. Combining them gives a negative fraction that is equal to the negative of the third tangent. This confirms their total sum is 180 degrees (\(\pi\)).

Exam Tip: Pay attention to signs. Since \( \tan(A+B) \) is negative and \( A, B > 0 \), the sum \( A+B \) lies in the second quadrant, which requires adding \( \pi \) to the inverse tangent formula.

Question 4. Prove that \( \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \pi \)
Answer: We calculate \( \tan^{-1} 1 = \frac{\pi}{4} \).
For the other two terms, since the product \( x \cdot y = 2 \cdot 3 = 6 > 1 \), we must use the sum formula for \( xy > 1 \):
\( \tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}\left(\frac{2+3}{1 - 6}\right) \)

\( \implies \tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}\left(\frac{5}{-5}\right) \)

\( \implies \tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}(-1) \)

\( \implies \tan^{-1} 2 + \tan^{-1} 3 = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
Now, we sum all three terms:
\( \text{Sum} = \frac{\pi}{4} + \frac{3\pi}{4} = \pi \).
Hence, proved.
In simple words: The first term is 45 degrees. Since the product of 2 and 3 is greater than 1, we add 180 degrees (\(\pi\)) when combining them, which reduces their sum to 135 degrees. Adding 45 and 135 degrees gives 180 degrees (\(\pi\)).

Exam Tip: Forgetting the \( \pi \) adjustment term when \( xy > 1 \) is the most common mistake students make on this classic exam question.

Question 5. Prove that \( \tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{x-y}{x+y}\right) = \frac{\pi}{4} \)
Answer: We rewrite the second term by dividing its numerator and denominator by \( y \):
\( \tan^{-1}\left(\frac{x-y}{x+y}\right) = \tan^{-1}\left(\frac{x/y - 1}{x/y + 1}\right) \).
Using the identity \( \tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1} a - \tan^{-1} b \) (with \( a = x/y \) and \( b = 1 \)):
\( \tan^{-1}\left(\frac{x/y - 1}{x/y + 1}\right) = \tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}(1) \).
Substituting this back into the original expression:
\( \text{LHS} = \tan^{-1}\left(\frac{x}{y}\right) - \left[\tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}(1)\right] \)

\( \implies \text{LHS} = \tan^{-1}(1) = \frac{\pi}{4} = \text{RHS} \).
Hence, proved.
In simple words: Divide the fraction in the second term by y. This lets you split it into the difference of two separate tangents, which cancels out the first term, leaving just the inverse tangent of 1 (which is 45 degrees).

Exam Tip: Recognizing the reverse application of the tangent subtraction formula can make difficult algebra problems much simpler.

Question 6. Write in the simplest form \( \cos\left[2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right] \)
Answer: Let \( x = \cos \theta \implies \theta = \cos^{-1} x \).
We substitute this into the inner expression:
\[ E = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \]
Using half-angle identities \( 1 - \cos \theta = 2\sin^2 \frac{\theta}{2} \) and \( 1 + \cos \theta = 2\cos^2 \frac{\theta}{2} \):
\[ E = \sqrt{\frac{2\sin^2 \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}}} = \sqrt{\tan^2 \frac{\theta}{2}} = \tan \frac{\theta}{2} \].
Now, we substitute this back into the main expression:
\[ \cos\left[2\tan^{-1}\left(\tan \frac{\theta}{2}\right)\right] = \cos\left[2 \cdot \frac{\theta}{2}\right] = \cos \theta \].
Since we defined \( \cos \theta = x \), the expression simplifies to:
\( x \).
The simplest form is \( x \).
In simple words: Substituting x with cos(theta) simplifies the square root fraction to tangent of half-theta. This cancels the inverse tangent, leaving cos(theta), which is just x.

Exam Tip: Always make sure to substitute your original variable back into your final simplified answer.

Page 18

Question 7. Solve \( \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4} \)
Answer: We combine the terms using the tangent sum identity:
\( \tan^{-1}\left( \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)} \right) = \frac{\pi}{4} \)
Taking the tangent on both sides:
\( \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan\left(\frac{\pi}{4}\right) \)

\( \implies \frac{(x^2 + x - 2) + (x^2 - x - 2)}{(x^2 - 4) - (x^2 - 1)} = 1 \)

\( \implies \frac{2x^2 - 4}{-3} = 1 \)

\( \implies 2x^2 - 4 = -3 \)

\( \implies 2x^2 = 1 \)

\( \implies x^2 = \frac{1}{2} \)

\( \implies x = \pm \frac{1}{\sqrt{2}} \).
Both values of \( x \) are valid and satisfy the original equation.
In simple words: Combine the tangents into a single fraction, simplify the algebraic terms, and solve for x to find the two answers.

Exam Tip: Be sure to write down each step of your fraction arithmetic clearly to avoid losing marks on exams.

Question 8. Solve \( \tan^{-1} 2x + \tan^{-1} 3x = \pi / 4 \)
Answer: We use the tangent addition formula to combine the terms on the left side:
\( \tan^{-1}\left(\frac{2x + 3x}{1 - 6x^2}\right) = \frac{\pi}{4} \)

\( \implies \frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right) \)

\( \implies \frac{5x}{1 - 6x^2} = 1 \)

\( \implies 5x = 1 - 6x^2 \)

\( \implies 6x^2 + 5x - 1 = 0 \).
Factoring the quadratic equation:
\( 6x^2 + 6x - x - 1 = 0 \)

\( \implies 6x(x+1) - 1(x+1) = 0 \)

\( \implies (6x-1)(x+1) = 0 \).
This gives the roots \( x = \frac{1}{6} \) or \( x = -1 \).

We test these solutions in the original equation:
If \( x = -1 \), the left side is negative and does not equal the positive right side, so we discard it.
Thus, the only valid solution is \( x = \frac{1}{6} \).
In simple words: Combine the tangent expressions, factor the resulting quadratic equation, and discard the negative root because it does not fit the equation's range.

Exam Tip: Remember to always check your final roots in the original equation to eliminate any invalid negative answers.