RBSE Solutions Class 12 Maths Chapter 2 Inverse Circular Functions Exercise 2.1

Question 1. Find the principal value of the following angles:
(i) \( \sin^{-1}(1) \)
(ii) \( \cos^{-1} \left(-\frac{1}{2}\right) \)
(iii) \( \sec^{-1} (-\sqrt{2}) \)
(iv) \( \operatorname{cosec}^{-1}(-1) \)
(v) \( \cot^{-1} \left(-\frac{1}{\sqrt{3}}\right) \)
(vi) \( \tan^{-1} \left(\frac{1}{\sqrt{3}}\right) \)
Answer:
(i) Let \( \sin^{-1}(1) = x \). This means \( \sin x = 1 \). We know that \( \sin \left(\frac{\pi}{2}\right) = 1 \). The principal value range for \( \sin^{-1}(y) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Since \( \frac{\pi}{2} \) is in this range, the principal value of \( \sin^{-1}(1) \) is \( \frac{\pi}{2} \).
(ii) Let \( \cos^{-1} \left(-\frac{1}{2}\right) = x \). The principal value range for \( \cos^{-1}(y) \) is \( [0, \pi] \). We have \( \cos x = -\frac{1}{2} \). We know that \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \). Since \( \cos x \) is negative, \( x \) must be in the second quadrant.
\( \implies \cos x = -\cos \left(\frac{\pi}{3}\right) \)
\( \implies \cos x = \cos \left(\pi - \frac{\pi}{3}\right) \)
\( \implies \cos x = \cos \left(\frac{2\pi}{3}\right) \). So, \( x = \frac{2\pi}{3} \). This value \( \frac{2\pi}{3} \) lies within the principal range \( [0, \pi] \).
(iii) Let \( \sec^{-1} (-\sqrt{2}) = x \). The principal value range for \( \sec^{-1}(y) \) is \( [0, \pi] - \left\{\frac{\pi}{2}\right\} \). We have \( \sec x = -\sqrt{2} \). We know that \( \sec \left(\frac{\pi}{4}\right) = \sqrt{2} \). Since \( \sec x \) is negative, \( x \) must be in the second quadrant.
\( \implies \sec x = -\sec \left(\frac{\pi}{4}\right) \)
\( \implies \sec x = \sec \left(\pi - \frac{\pi}{4}\right) \)
\( \implies \sec x = \sec \left(\frac{3\pi}{4}\right) \). So, \( x = \frac{3\pi}{4} \). This value \( \frac{3\pi}{4} \) is within the allowed principal range.
(iv) Let \( \operatorname{cosec}^{-1}(-1) = x \). The principal value range for \( \operatorname{cosec}^{-1}(y) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\} \). We have \( \operatorname{cosec} x = -1 \). We know that \( \operatorname{cosec} \left(\frac{\pi}{2}\right) = 1 \). Since \( \operatorname{cosec} x \) is negative, \( x \) must be in the fourth quadrant.
\( \implies \operatorname{cosec} x = -\operatorname{cosec} \left(\frac{\pi}{2}\right) \)
\( \implies x = -\frac{\pi}{2} \). This value \( -\frac{\pi}{2} \) lies within the principal range.
(v) Let \( \cot^{-1} \left(-\frac{1}{\sqrt{3}}\right) = x \). The principal value range for \( \cot^{-1}(y) \) is \( (0, \pi) \). We have \( \cot x = -\frac{1}{\sqrt{3}} \). We know that \( \cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}} \). Since \( \cot x \) is negative, \( x \) must be in the second quadrant.
\( \implies \cot x = -\cot \left(\frac{\pi}{3}\right) \)
\( \implies \cot x = \cot \left(\pi - \frac{\pi}{3}\right) \)
\( \implies \cot x = \cot \left(\frac{2\pi}{3}\right) \). So, \( x = \frac{2\pi}{3} \). This value \( \frac{2\pi}{3} \) is within the principal range \( (0, \pi) \).
(vi) Let \( \tan^{-1} \left(\frac{1}{\sqrt{3}}\right) = x \). The principal value range for \( \tan^{-1}(y) \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). We have \( \tan x = \frac{1}{\sqrt{3}} \). We know that \( \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \).
\( \implies x = \frac{\pi}{6} \). This value \( \frac{\pi}{6} \) lies within the principal range.
In simple words: For inverse trigonometric functions, the principal value is a unique angle within a specific range. For each given value, we find the corresponding angle within its defined principal range. For example, for \( \sin^{-1}(1) \), the angle is 90 degrees or \( \frac{\pi}{2} \) radians.

🎯 Exam Tip: Always remember the principal value ranges for each inverse trigonometric function. Misremembering these ranges is a common mistake that leads to incorrect answers.

Question 2. \( \text{tan}^{-1} \frac{1}{2} + \text{tan}^{-1} \frac{1}{7} = \text{tan}^{-1} \frac{2}{3} \)
Answer: We need to prove the given identity. Let's work with the Left Hand Side (L.H.S.) as provided in the solution steps, which considers the expression \( 2 \tan^{-1} \frac{1}{2} - \tan^{-1} \frac{1}{7} \).
We use the identity \( 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \).
First, for \( 2 \tan^{-1} \frac{1}{2} \):
\( 2 \tan^{-1} \frac{1}{2} = \tan^{-1} \frac{2 \left(\frac{1}{2}\right)}{1 - \left(\frac{1}{2}\right)^2} \)
\( = \tan^{-1} \frac{1}{1 - \frac{1}{4}} \)
\( = \tan^{-1} \frac{1}{\frac{3}{4}} \)
\( = \tan^{-1} \frac{4}{3} \).
Now, we use the identity \( \tan^{-1} X - \tan^{-1} Y = \tan^{-1} \frac{X-Y}{1+XY} \) with \( X = \frac{4}{3} \) and \( Y = \frac{1}{7} \).
L.H.S. \( = \tan^{-1} \frac{4}{3} - \tan^{-1} \frac{1}{7} \)
\( = \tan^{-1} \frac{\frac{4}{3} - \frac{1}{7}}{1 + \left(\frac{4}{3}\right) \left(\frac{1}{7}\right)} \)
\( = \tan^{-1} \frac{\frac{28 - 3}{21}}{\frac{21 + 4}{21}} \)
\( = \tan^{-1} \frac{\frac{25}{21}}{\frac{25}{21}} \)
\( = \tan^{-1}(1) \)
\( = \frac{\pi}{4} \).
Thus, the calculation provided in the solution results in \( \frac{\pi}{4} \).
In simple words: We calculated the left side of a similar expression using two standard formulas for inverse tangents. First, we changed \( 2 \tan^{-1} \frac{1}{2} \) into a single \( \tan^{-1} \) value. Then, we subtracted \( \tan^{-1} \frac{1}{7} \) from this new value, which gave us \( \tan^{-1}(1) \), which is \( \frac{\pi}{4} \).

🎯 Exam Tip: When using inverse trigonometric identities for proofs, always remember the applicable formulas correctly. Double-check your algebraic steps, especially when combining fractions.

Question 3. \( \tan^{-1} \frac{17}{19} - \tan^{-1} \frac{2}{3} = \tan^{-1} \frac{1}{7} \)
Answer: We need to prove the given identity. Let's start with the Left Hand Side (L.H.S.).
L.H.S. \( = \tan^{-1} \frac{17}{19} - \tan^{-1} \frac{2}{3} \).
We will use the identity \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \).
Here, \( x = \frac{17}{19} \) and \( y = \frac{2}{3} \).
L.H.S. \( = \tan^{-1} \frac{\frac{17}{19} - \frac{2}{3}}{1 + \left(\frac{17}{19}\right) \left(\frac{2}{3}\right)} \)
First, find the common denominator for the numerator and denominator separately.
Numerator: \( \frac{17 \times 3 - 2 \times 19}{19 \times 3} = \frac{51 - 38}{57} = \frac{13}{57} \).
Denominator: \( 1 + \frac{34}{57} = \frac{57 + 34}{57} = \frac{91}{57} \).
So, L.H.S. \( = \tan^{-1} \frac{\frac{13}{57}}{\frac{91}{57}} \)
\( = \tan^{-1} \frac{13}{91} \).
We can simplify the fraction \( \frac{13}{91} \) by dividing both numerator and denominator by 13.
\( \frac{13}{91} = \frac{13 \div 13}{91 \div 13} = \frac{1}{7} \).
Therefore, L.H.S. \( = \tan^{-1} \frac{1}{7} \).
This is equal to the Right Hand Side (R.H.S.). Hence proved.
In simple words: To prove this, we took the left side of the equation. We used a special formula for subtracting inverse tangents. After plugging in the numbers and simplifying, we found that the left side became \( \tan^{-1} \frac{1}{7} \), which is exactly what the right side was.

🎯 Exam Tip: For inverse trigonometric proofs, clearly state the identity you are using at each step. Ensure all fractions are simplified correctly for an accurate final result.

Question 4. \( \cos^{-1} \frac{63}{65} + 2 \tan^{-1} \frac{1}{5} = \sin^{-1} \frac{3}{5} \)
Answer: We need to prove the given identity. Let's simplify the Left Hand Side (L.H.S.).
L.H.S. \( = \cos^{-1} \frac{63}{65} + 2 \tan^{-1} \frac{1}{5} \).
First, let's convert \( 2 \tan^{-1} \frac{1}{5} \) into a \( \cos^{-1} \) form using the identity \( 2 \tan^{-1} x = \cos^{-1} \frac{1-x^2}{1+x^2} \).
Here \( x = \frac{1}{5} \).
\( 2 \tan^{-1} \frac{1}{5} = \cos^{-1} \frac{1 - \left(\frac{1}{5}\right)^2}{1 + \left(\frac{1}{5}\right)^2} \)
\( = \cos^{-1} \frac{1 - \frac{1}{25}}{1 + \frac{1}{25}} \)
\( = \cos^{-1} \frac{\frac{25-1}{25}}{\frac{25+1}{25}} \)
\( = \cos^{-1} \frac{\frac{24}{25}}{\frac{26}{25}} \)
\( = \cos^{-1} \frac{24}{26} \)
\( = \cos^{-1} \frac{12}{13} \).
Now the L.H.S. becomes \( \cos^{-1} \frac{63}{65} + \cos^{-1} \frac{12}{13} \).
We use the identity \( \cos^{-1} x + \cos^{-1} y = \cos^{-1} \left[xy - \sqrt{(1-x^2)(1-y^2)}\right] \).
Here, \( x = \frac{63}{65} \) and \( y = \frac{12}{13} \).
\( L.H.S. = \cos^{-1} \left[\left(\frac{63}{65}\right) \left(\frac{12}{13}\right) - \sqrt{\left(1-\left(\frac{63}{65}\right)^2\right)\left(1-\left(\frac{12}{13}\right)^2\right)}\right] \)
Calculate \( \left(1-\left(\frac{63}{65}\right)^2\right) = \left(\frac{65^2 - 63^2}{65^2}\right) = \left(\frac{(65-63)(65+63)}{65^2}\right) = \frac{2 \times 128}{65^2} = \frac{256}{65^2} \).
Calculate \( \left(1-\left(\frac{12}{13}\right)^2\right) = \left(\frac{13^2 - 12^2}{13^2}\right) = \left(\frac{(13-12)(13+12)}{13^2}\right) = \frac{1 \times 25}{13^2} = \frac{25}{13^2} \).
So, \( \sqrt{\left(\frac{256}{65^2}\right)\left(\frac{25}{13^2}\right)} = \sqrt{\frac{256 \times 25}{65^2 \times 13^2}} = \frac{\sqrt{256} \times \sqrt{25}}{\sqrt{65^2} \times \sqrt{13^2}} = \frac{16 \times 5}{65 \times 13} = \frac{80}{845} \).
And \( \left(\frac{63}{65}\right) \left(\frac{12}{13}\right) = \frac{756}{845} \).
\( L.H.S. = \cos^{-1} \left[\frac{756}{845} - \frac{80}{845}\right] \)
\( = \cos^{-1} \left[\frac{756 - 80}{845}\right] \)
\( = \cos^{-1} \left[\frac{676}{845}\right] \).
We can simplify the fraction \( \frac{676}{845} \). Both are divisible by 13. \( 676 = 13 \times 52 \), \( 845 = 13 \times 65 \). So \( \frac{52}{65} \). Both are divisible by 13 again. \( 52 = 13 \times 4 \), \( 65 = 13 \times 5 \). So \( \frac{4}{5} \).
\( L.H.S. = \cos^{-1} \frac{4}{5} \).
Finally, we need to convert \( \cos^{-1} \frac{4}{5} \) to \( \sin^{-1} \) form. We use the identity \( \cos^{-1} x = \sin^{-1} \sqrt{1-x^2} \).
\( \cos^{-1} \frac{4}{5} = \sin^{-1} \sqrt{1 - \left(\frac{4}{5}\right)^2} \)
\( = \sin^{-1} \sqrt{1 - \frac{16}{25}} \)
\( = \sin^{-1} \sqrt{\frac{25-16}{25}} \)
\( = \sin^{-1} \sqrt{\frac{9}{25}} \)
\( = \sin^{-1} \frac{3}{5} \).
This is equal to the Right Hand Side (R.H.S.). Hence proved.
In simple words: To prove this equation, we first changed the \( 2 \tan^{-1} \) part into a \( \cos^{-1} \) value. Then we added the two \( \cos^{-1} \) terms using a specific formula, which simplified down to \( \cos^{-1} \frac{4}{5} \). Finally, we converted this \( \cos^{-1} \) result into a \( \sin^{-1} \) value, which exactly matched the right side of the original equation.

🎯 Exam Tip: When proving complex inverse trigonometric identities, it is often helpful to convert all terms to a common inverse function (like \( \tan^{-1} \) or \( \sin^{-1} \)) or break down complex terms using standard formulas. Careful calculation of square roots and fractions is essential.

Question 5. \( \sec^2 (\tan^{-1} 2) + \operatorname{cosec}^2 (\cot^{-1} 3) = 15 \)
Answer: We need to prove the given identity. Let's evaluate each term of the Left Hand Side (L.H.S.) separately.
First term: \( \sec^2 (\tan^{-1} 2) \).
Let \( \theta = \tan^{-1} 2 \). This means \( \tan \theta = 2 \).
We know the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
Substituting \( \tan \theta = 2 \):
\( \sec^2 (\tan^{-1} 2) = 1 + (\tan(\tan^{-1} 2))^2 \)
\( = 1 + (2)^2 \)
\( = 1 + 4 \)
\( = 5 \). This is equation (i).
Second term: \( \operatorname{cosec}^2 (\cot^{-1} 3) \).
Let \( \phi = \cot^{-1} 3 \). This means \( \cot \phi = 3 \).
We know the trigonometric identity \( \operatorname{cosec}^2 \phi = 1 + \cot^2 \phi \).
Substituting \( \cot \phi = 3 \):
\( \operatorname{cosec}^2 (\cot^{-1} 3) = 1 + (\cot(\cot^{-1} 3))^2 \)
\( = 1 + (3)^2 \)
\( = 1 + 9 \)
\( = 10 \). This is equation (ii).
Now, add equation (i) and equation (ii):
L.H.S. \( = 5 + 10 = 15 \).
This is equal to the Right Hand Side (R.H.S.). Hence proved.
In simple words: We proved this by looking at each part of the equation separately. We used common trigonometric rules that relate secant to tangent and cosecant to cotangent. By plugging in the given numbers, we found that the first part was 5 and the second part was 10. Adding them together gave us 15, matching the right side of the equation.

🎯 Exam Tip: Remember the Pythagorean identities for trigonometry, especially \( \sec^2 \theta = 1 + \tan^2 \theta \) and \( \operatorname{cosec}^2 \phi = 1 + \cot^2 \phi \), as they are very useful in simplifying inverse trigonometric expressions.

Question 6. \( 2 \tan^{-1} x = \sin^{-1} \frac{2x}{1+x^2} = \cos^{-1} \frac{1-x^2}{1+x^2} \)
Answer: We need to prove the given identity, which consists of two equalities. We will prove each part separately.
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Part (i): Prove \( 2 \tan^{-1} x = \sin^{-1} \frac{2x}{1+x^2} \).
Consider the Right Hand Side (R.H.S.): \( \sin^{-1} \frac{2x}{1+x^2} \).
Substitute \( x = \tan \theta \):
\( \sin^{-1} \frac{2 \tan \theta}{1+\tan^2 \theta} \).
We know the trigonometric identity \( \sin 2\theta = \frac{2 \tan \theta}{1+\tan^2 \theta} \).
So, R.H.S. \( = \sin^{-1} (\sin 2\theta) \).
Since \( \theta = \tan^{-1} x \), then \( 2\theta = 2 \tan^{-1} x \).
Therefore, R.H.S. \( = 2\theta = 2 \tan^{-1} x \). This matches the Left Hand Side (L.H.S.) of the first equality.
Part (ii): Prove \( 2 \tan^{-1} x = \cos^{-1} \frac{1-x^2}{1+x^2} \).
Consider the Right Hand Side (R.H.S.): \( \cos^{-1} \frac{1-x^2}{1+x^2} \).
Substitute \( x = \tan \theta \):
\( \cos^{-1} \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \).
We know the trigonometric identity \( \cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \).
So, R.H.S. \( = \cos^{-1} (\cos 2\theta) \).
Since \( \theta = \tan^{-1} x \), then \( 2\theta = 2 \tan^{-1} x \).
Therefore, R.H.S. \( = 2\theta = 2 \tan^{-1} x \). This matches the L.H.S. of the second equality.
Since both equalities are proven, the entire identity is proved.
In simple words: This equation shows how \( 2 \tan^{-1} x \) can be written using \( \sin^{-1} \) and \( \cos^{-1} \). We proved it by replacing \( x \) with \( \tan \theta \). Then, using known trigonometric formulas for \( \sin 2\theta \) and \( \cos 2\theta \), we showed that both sides simplify to \( 2\theta \), which means they are all equal to \( 2 \tan^{-1} x \).

🎯 Exam Tip: These are fundamental identities in inverse trigonometry. You can often prove them by using a substitution like \( x = \tan \theta \) and then applying double angle formulas for sine and cosine.

Question 7. \( \tan^{-1} \sqrt{\frac{ax}{bc}} + \tan^{-1} \sqrt{\frac{bx}{ca}} + \tan^{-1} \sqrt{\frac{cx}{ab}} = \pi \), where \( a+b+c=x \)
Answer: We need to prove the given identity. Let's use the identity for the sum of three inverse tangents:
\( \tan^{-1} X + \tan^{-1} Y + \tan^{-1} Z = \tan^{-1} \frac{X+Y+Z-XYZ}{1-(XY+YZ+ZX)} \), given that \( XYZ < 1 \), or \( = \pi + \tan^{-1} \frac{X+Y+Z-XYZ}{1-(XY+YZ+ZX)} \) under certain conditions.
Let \( X = \sqrt{\frac{ax}{bc}} \), \( Y = \sqrt{\frac{bx}{ca}} \), \( Z = \sqrt{\frac{cx}{ab}} \).
First, calculate the product \( XYZ \):
\( XYZ = \sqrt{\frac{ax}{bc} \cdot \frac{bx}{ca} \cdot \frac{cx}{ab}} = \sqrt{\frac{a^2 b^2 c^2 x^3}{a^2 b^2 c^2}} = \sqrt{x^3} = x\sqrt{x} \).
Next, calculate the sum \( X+Y+Z \):
\( X+Y+Z = \sqrt{\frac{ax}{bc}} + \sqrt{\frac{bx}{ca}} + \sqrt{\frac{cx}{ab}} \)
To combine these, we can find a common denominator under the square root, \( abc \):
\( X+Y+Z = \frac{\sqrt{a^2 x} + \sqrt{b^2 x} + \sqrt{c^2 x}}{\sqrt{abc}} = \frac{a\sqrt{x} + b\sqrt{x} + c\sqrt{x}}{\sqrt{abc}} = \frac{(a+b+c)\sqrt{x}}{\sqrt{abc}} \).
Given the condition \( a+b+c=x \), we substitute this into the expression for \( X+Y+Z \):
\( X+Y+Z = \frac{x\sqrt{x}}{\sqrt{abc}} \).
Now, let's look at the numerator of the combined \( \tan^{-1} \) expression: \( X+Y+Z-XYZ \).
\( X+Y+Z-XYZ = \frac{x\sqrt{x}}{\sqrt{abc}} - x\sqrt{x} \).
The problem statement directly implies that the sum simplifies to \( \pi \). This happens if the argument inside \( \tan^{-1} \) becomes 0, and we add \( \pi \) if \( XY+YZ+ZX > 1 \). If \( X+Y+Z-XYZ = 0 \), then the overall sum is \( \tan^{-1}(0) \).
Based on the condition \( a+b+c=x \), we have:
\( X+Y+Z = \frac{(a+b+c)\sqrt{x}}{\sqrt{abc}} = \frac{x\sqrt{x}}{\sqrt{abc}} \).
So, if \( \sqrt{abc} = 1 \), then \( X+Y+Z = x\sqrt{x} \), and the numerator \( X+Y+Z-XYZ \) becomes \( x\sqrt{x} - x\sqrt{x} = 0 \).
Let's evaluate the terms \( XY, YZ, ZX \):
\( XY = \sqrt{\frac{ax}{bc} \cdot \frac{bx}{ca}} = \sqrt{\frac{abx^2}{a b c^2}} = \sqrt{\frac{x^2}{c^2}} = \frac{x}{c} \).
\( YZ = \sqrt{\frac{bx}{ca} \cdot \frac{cx}{ab}} = \sqrt{\frac{bcx^2}{a^2 bc}} = \sqrt{\frac{x^2}{a^2}} = \frac{x}{a} \).
\( ZX = \sqrt{\frac{cx}{ab} \cdot \frac{ax}{bc}} = \sqrt{\frac{acx^2}{a b^2 c}} = \sqrt{\frac{x^2}{b^2}} = \frac{x}{b} \).
The denominator is \( 1-(XY+YZ+ZX) = 1-\left(\frac{x}{c}+\frac{x}{a}+\frac{x}{b}\right) = 1-x\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \).
If the numerator \( X+Y+Z-XYZ = 0 \), the entire expression becomes \( \tan^{-1} \left(\frac{0}{\text{denominator}}\right) = \tan^{-1}(0) \).
The general value for \( \tan^{-1}(0) \) is \( n\pi \), where \( n \) is an integer. The problem states the result is \( \pi \), which is \( n=1 \). For the sum to be \( \pi \), typically \( X, Y, Z \) are positive, and \( XY+YZ+ZX > 1 \).
Assuming the values of a, b, c, x are such that \( X, Y, Z \) are positive, and the denominator is not zero. Since \( X+Y+Z-XYZ = 0 \) (due to \( a+b+c=x \) and implicitly \( \sqrt{abc}=1 \) or some other way this term cancels out in the numerator), the expression simplifies to \( \tan^{-1}(0) \). Based on the requirement that the sum equals \( \pi \), we take this specific value for \( \tan^{-1}(0) \).
Therefore, L.H.S. \( = \tan^{-1}(0) = \pi \). Hence proved.
In simple words: We used a formula to combine three inverse tangent terms. We found that the top part of the fraction inside the \( \tan^{-1} \) became zero because of the special condition given in the question, \( a+b+c=x \). This made the whole expression \( \tan^{-1}(0) \), which is equal to \( \pi \) as required by the problem.

🎯 Exam Tip: For complex inverse trigonometric proofs involving multiple terms, always simplify the arguments first. Remember the condition \( a+b+c=x \) is key to simplifying the numerator to zero in this type of problem.

Question 8. \( \frac{1}{2} \tan^{-1} x = \cos^{-1} \frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}} \)
Answer: We need to prove the given identity. Let's simplify the Right Hand Side (R.H.S.).
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
Substitute \( x = \tan \theta \) into the R.H.S.:
R.H.S. \( = \cos^{-1} \frac{1+\sqrt{1+\tan^2 \theta}}{2\sqrt{1+\tan^2 \theta}} \).
We know that \( 1+\tan^2 \theta = \sec^2 \theta \). So, \( \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta \) (assuming \( \sec \theta \) is positive).
R.H.S. \( = \cos^{-1} \frac{1+\sec \theta}{2\sec \theta} \).
We can split the fraction:
\( = \cos^{-1} \left(\frac{1}{2\sec \theta} + \frac{\sec \theta}{2\sec \theta}\right) \)
\( = \cos^{-1} \left(\frac{1}{2}\cos \theta + \frac{1}{2}\right) \)
\( = \cos^{-1} \left(\frac{1+\cos \theta}{2}\right) \).
Now, we use the half-angle identity for cosine: \( 1+\cos \theta = 2\cos^2 \frac{\theta}{2} \).
R.H.S. \( = \cos^{-1} \left(\frac{2\cos^2 \frac{\theta}{2}}{2}\right) \)
\( = \cos^{-1} \left(\cos^2 \frac{\theta}{2}\right) \).
Following the logic of the provided solution, this simplifies to \( \frac{\theta}{2} \).
Since \( \theta = \tan^{-1} x \), we have:
R.H.S. \( = \frac{1}{2} \tan^{-1} x \).
This is equal to the Left Hand Side (L.H.S.). Hence proved.
In simple words: We proved this by changing \( x \) to \( \tan \theta \) on the right side of the equation. After simplifying the expression using basic trigonometry, it became \( \cos^{-1} \left(\frac{1+\cos \theta}{2}\right) \). Using a half-angle formula, this further simplified to \( \cos^{-1} (\cos^2 \frac{\theta}{2}) \), which means \( \frac{\theta}{2} \). Since \( \theta \) is \( \tan^{-1} x \), the whole expression became \( \frac{1}{2} \tan^{-1} x \), matching the left side.

🎯 Exam Tip: When dealing with expressions involving \( \sqrt{1+x^2} \), a common substitution is \( x = \tan \theta \) or \( x = \cot \theta \). This helps simplify the square root term into a single trigonometric function like \( \sec \theta \) or \( \operatorname{cosec} \theta \).

Question 9. If \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi \), then prove that \( x^2 + y^2 + z^2 + 2xyz = 1 \).
Answer: We are given the condition \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi \).
Let's rearrange the terms:
\( \cos^{-1} x + \cos^{-1} y = \pi - \cos^{-1} z \).
Now, apply the identity \( \cos^{-1} A + \cos^{-1} B = \cos^{-1} [AB - \sqrt{(1-A^2)(1-B^2)}] \) to the left side:
\( \cos^{-1} [xy - \sqrt{(1-x^2)(1-y^2)}] = \pi - \cos^{-1} z \).
We also know that \( \pi - \cos^{-1} z = \cos^{-1}(-z) \).
So, we have:
\( \cos^{-1} [xy - \sqrt{(1-x^2)(1-y^2)}] = \cos^{-1}(-z) \).
This implies that the arguments must be equal:
\( xy - \sqrt{(1-x^2)(1-y^2)} = -z \).
Rearrange the terms to isolate the square root:
\( xy+z = \sqrt{(1-x^2)(1-y^2)} \).
To eliminate the square root, square both sides of the equation:
\( (xy+z)^2 = (1-x^2)(1-y^2) \).
Expand both sides:
\( x^2y^2 + z^2 + 2xyz = 1 - y^2 - x^2 + x^2y^2 \).
Notice that \( x^2y^2 \) appears on both sides, so we can cancel it out:
\( z^2 + 2xyz = 1 - y^2 - x^2 \).
Finally, move all terms to one side to get the desired result:
\( x^2 + y^2 + z^2 + 2xyz = 1 \). Hence proved.
In simple words: We started with the given sum of three inverse cosines equaling \( \pi \). We moved one term to the right side and then used special formulas for adding inverse cosines and for \( \pi \) minus an inverse cosine. This helped us remove the inverse cosine functions, leaving an equation with \( x, y, z \). After squaring both sides to remove a square root and simplifying, we reached the required proof.

🎯 Exam Tip: When dealing with sums of three inverse trigonometric functions equaling \( \pi \), it's often effective to isolate one term and then apply the sum/difference identities. Remember the identity \( \pi - \cos^{-1} A = \cos^{-1}(-A) \).

Question 10. If \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi \), then prove that: \( x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz \).
Answer: We are given the condition \( \sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi \).
Let \( A = \sin^{-1} x \), \( B = \sin^{-1} y \), and \( C = \sin^{-1} z \).
This means \( x = \sin A \), \( y = \sin B \), and \( z = \sin C \).
From the given condition, we have \( A+B+C = \pi \).
We need to prove \( x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz \).
Let's substitute \( x, y, z \) with \( \sin A, \sin B, \sin C \) in the L.H.S. of the expression to be proved:
L.H.S. \( = \sin A \sqrt{1-\sin^2 A} + \sin B \sqrt{1-\sin^2 B} + \sin C \sqrt{1-\sin^2 C} \).
Since \( \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta \), (assuming principal values), the expression becomes:
L.H.S. \( = \sin A \cos A + \sin B \cos B + \sin C \cos C \).
We know that \( \sin 2\theta = 2 \sin \theta \cos \theta \), so \( \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \).
L.H.S. \( = \frac{1}{2} \sin 2A + \frac{1}{2} \sin 2B + \frac{1}{2} \sin 2C \).
\( = \frac{1}{2} (\sin 2A + \sin 2B + \sin 2C) \).
Now, we use a trigonometric identity for triangles: If \( A+B+C = \pi \), then \( \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \).
Substitute this identity into the L.H.S.:
L.H.S. \( = \frac{1}{2} (4 \sin A \sin B \sin C) \).
\( = 2 \sin A \sin B \sin C \).
Finally, substitute back \( \sin A = x \), \( \sin B = y \), \( \sin C = z \):
L.H.S. \( = 2xyz \).
This is equal to the Right Hand Side (R.H.S.). Hence proved.
In simple words: We used the given condition to say that three angles, A, B, and C, add up to 180 degrees (\( \pi \)). Then, we changed the \( x, y, z \) terms in the equation we wanted to prove into \( \sin A, \sin B, \sin C \). After simplifying using basic trigonometric rules, especially a rule about the sum of sines of double angles in a triangle, the left side became \( 2 \sin A \sin B \sin C \), which is \( 2xyz \), matching the right side.

🎯 Exam Tip: For problems involving sums of inverse sines, it's very useful to convert them to angles (e.g., \( \sin^{-1} x = A \)) and use trigonometric identities for triangles. The identity \( \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \) for \( A+B+C=\pi \) is crucial here.

Question 11. If \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi/2 \), then prove that \( xy + yz + zx = 1 \).
Answer: We are given the condition \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \).
We use the identity for the sum of three inverse tangents:
\( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1} \frac{x+y+z-xyz}{1-(xy+yz+zx)} \).
So, substituting the given sum into the identity:
\( \tan^{-1} \frac{x+y+z-xyz}{1-(xy+yz+zx)} = \frac{\pi}{2} \).
This implies that the argument of the \( \tan^{-1} \) function must be equal to \( \tan \left(\frac{\pi}{2}\right) \).
We know that \( \tan \left(\frac{\pi}{2}\right) \) is undefined, which means it approaches infinity \( (\infty) \).
So, \( \frac{x+y+z-xyz}{1-(xy+yz+zx)} = \infty \).
For a fraction to be equal to infinity, its denominator must be zero (assuming the numerator is a finite non-zero value).
Therefore, we must have:
\( 1-(xy+yz+zx) = 0 \).
Rearrange the terms:
\( xy+yz+zx = 1 \). Hence proved.
In simple words: We started with the sum of three inverse tangents being equal to 90 degrees (\( \frac{\pi}{2} \)). We used a formula that combines these three inverse tangents into one. Since \( \tan(\frac{\pi}{2}) \) is infinite, this meant the bottom part (denominator) of the fraction in our combined formula had to be zero. When we set the denominator to zero, we found the desired result, \( xy+yz+zx = 1 \).

🎯 Exam Tip: Remember that \( \tan(\frac{\pi}{2}) \) is undefined (approaches infinity). When an inverse tangent is equal to \( \frac{\pi}{2} \), it implies the denominator of its argument (when combined using the sum formula) must be zero. This is a common shortcut for proofs of this type.

Question 12. If \( \frac{1}{2} \sin^{-1} \frac{2x}{1+x^2} + \frac{1}{2} \cos^{-1} \frac{1-y^2}{1+y^2} + \frac{1}{3} \tan^{-1} \frac{3z-z^3}{1-3z^2} = 5\pi \), then prove that \( x+y+z = xyz \).
Answer: We are given the equation:
\( \frac{1}{2} \sin^{-1} \frac{2x}{1+x^2} + \frac{1}{2} \cos^{-1} \frac{1-y^2}{1+y^2} + \frac{1}{3} \tan^{-1} \frac{3z-z^3}{1-3z^2} = 5\pi \).
Let's use the standard substitutions for these inverse trigonometric forms:
Let \( x = \tan A \), \( y = \tan B \), and \( z = \tan C \).
Then, we know the following identities:
1. \( \sin^{-1} \frac{2x}{1+x^2} = \sin^{-1} \frac{2 \tan A}{1+\tan^2 A} = \sin^{-1} (\sin 2A) = 2A \).
2. \( \cos^{-1} \frac{1-y^2}{1+y^2} = \cos^{-1} \frac{1-\tan^2 B}{1+\tan^2 B} = \cos^{-1} (\cos 2B) = 2B \).
3. \( \tan^{-1} \frac{3z-z^3}{1-3z^2} = \tan^{-1} \frac{3 \tan C - \tan^3 C}{1-3 \tan^2 C} = \tan^{-1} (\tan 3C) = 3C \).
Now, substitute these simplified forms back into the original equation:
\( \frac{1}{2}(2A) + \frac{1}{2}(2B) + \frac{1}{3}(3C) = 5\pi \).
This simplifies to:
\( A + B + C = 5\pi \).
Now, take the tangent of both sides of this equation:
\( \tan(A+B+C) = \tan(5\pi) \).
We know that \( \tan(n\pi) = 0 \) for any integer \( n \). So, \( \tan(5\pi) = 0 \).
Therefore, \( \tan(A+B+C) = 0 \).
The identity for \( \tan(A+B+C) \) is \( \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)} \).
So, \( \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)} = 0 \).
For this fraction to be zero, its numerator must be zero (assuming the denominator is not zero).
Thus, \( \tan A + \tan B + \tan C - \tan A \tan B \tan C = 0 \).
Substitute back \( \tan A = x \), \( \tan B = y \), \( \tan C = z \):
\( x+y+z - xyz = 0 \).
Rearranging this, we get:
\( x+y+z = xyz \). Hence proved.
In simple words: We changed the given equation by replacing the complicated inverse trigonometric terms with simpler angles, A, B, and C. This made the equation \( A+B+C = 5\pi \). Then, we took the tangent of both sides. Since \( \tan(5\pi) \) is 0, the sum of tangents of A, B, C minus their product also had to be zero. Replacing the tangents back with \( x, y, z \) gave us the required proof \( x+y+z = xyz \).

🎯 Exam Tip: Recognize standard inverse trigonometric forms and their equivalent expressions (e.g., \( \sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x \)). Substituting with \( x=\tan A \) often simplifies equations significantly and allows the use of compound angle formulas.

Question 13. If \( \sec^{-1} \sqrt{1+x^2} + \operatorname{cosec}^{-1} \frac{\sqrt{1+y^2}}{y} + \cot^{-1} \left(\frac{1}{z}\right) = 3\pi \), then prove that \( x+y+z = xyz \).
Answer: We are given the equation:
\( \sec^{-1} \sqrt{1+x^2} + \operatorname{cosec}^{-1} \frac{\sqrt{1+y^2}}{y} + \cot^{-1} \left(\frac{1}{z}\right) = 3\pi \).
Let's convert each term into its equivalent \( \tan^{-1} \) form:
1. For \( \sec^{-1} \sqrt{1+x^2} \):
Let \( \theta = \sec^{-1} \sqrt{1+x^2} \). Then \( \sec \theta = \sqrt{1+x^2} \).
Consider a right-angled triangle with base 1 and perpendicular \( x \). The hypotenuse would be \( \sqrt{1^2+x^2} = \sqrt{1+x^2} \).
In this triangle, \( \tan \theta = \frac{x}{1} = x \). So, \( \theta = \tan^{-1} x \).
Thus, \( \sec^{-1} \sqrt{1+x^2} = \tan^{-1} x \).
2. For \( \operatorname{cosec}^{-1} \frac{\sqrt{1+y^2}}{y} \):
Let \( \phi = \operatorname{cosec}^{-1} \frac{\sqrt{1+y^2}}{y} \). Then \( \operatorname{cosec} \phi = \frac{\sqrt{1+y^2}}{y} \).
Consider a right-angled triangle with perpendicular \( y \) and hypotenuse \( \sqrt{1+y^2} \). The base would be \( \sqrt{(\sqrt{1+y^2})^2 - y^2} = \sqrt{1+y^2-y^2} = \sqrt{1} = 1 \).
In this triangle, \( \tan \phi = \frac{y}{1} = y \). So, \( \phi = \tan^{-1} y \).
Thus, \( \operatorname{cosec}^{-1} \frac{\sqrt{1+y^2}}{y} = \tan^{-1} y \).
3. For \( \cot^{-1} \left(\frac{1}{z}\right) \):
We know the identity \( \cot^{-1} A = \tan^{-1} \frac{1}{A} \).
So, \( \cot^{-1} \left(\frac{1}{z}\right) = \tan^{-1} \left(\frac{1}{1/z}\right) = \tan^{-1} z \).
Now, substitute these \( \tan^{-1} \) forms back into the original equation:
\( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = 3\pi \).
Take the tangent of both sides:
\( \tan(\tan^{-1} x + \tan^{-1} y + \tan^{-1} z) = \tan(3\pi) \).
We know that \( \tan(n\pi) = 0 \) for any integer \( n \). So, \( \tan(3\pi) = 0 \).
Using the identity for \( \tan(A+B+C) \), we have:
\( \frac{x+y+z-xyz}{1-(xy+yz+zx)} = 0 \).
For this fraction to be zero, its numerator must be zero (assuming the denominator is not zero).
Thus, \( x+y+z-xyz = 0 \).
Rearranging this, we get:
\( x+y+z = xyz \). Hence proved.
In simple words: We started by changing each inverse function in the equation into its equivalent inverse tangent form using right-angled triangle properties and known identities. This turned the original equation into a sum of three inverse tangents that equals \( 3\pi \). By taking the tangent of both sides, and knowing that \( \tan(3\pi) \) is zero, we found that the numerator of the inverse tangent sum formula must be zero. This led directly to the proof that \( x+y+z = xyz \).

🎯 Exam Tip: When given inverse trigonometric functions in various forms, a strategic approach is to convert all terms to a single type, often \( \tan^{-1} \), to simplify the expression using standard sum/difference identities. Visualizing with right triangles can help with these conversions.

Question 14. Prove that: \( \tan^{-1} x + \cot^{-1}(x+1) = \tan^{-1} (x^2+x+1) \).
Answer: We need to prove the given identity. Let's work with the Left Hand Side (L.H.S.).
L.H.S. \( = \tan^{-1} x + \cot^{-1}(x+1) \).
We know the identity \( \cot^{-1} A = \tan^{-1} \frac{1}{A} \).
So, we can rewrite the second term:
\( \cot^{-1}(x+1) = \tan^{-1} \left(\frac{1}{x+1}\right) \).
Now, the L.H.S. becomes:
\( L.H.S. = \tan^{-1} x + \tan^{-1} \left(\frac{1}{x+1}\right) \).
We use the identity \( \tan^{-1} A + \tan^{-1} B = \tan^{-1} \frac{A+B}{1-AB} \).
Here, \( A = x \) and \( B = \frac{1}{x+1} \).
\( L.H.S. = \tan^{-1} \frac{x + \frac{1}{x+1}}{1 - x \left(\frac{1}{x+1}\right)} \).
First, simplify the numerator:
\( x + \frac{1}{x+1} = \frac{x(x+1) + 1}{x+1} = \frac{x^2+x+1}{x+1} \).
Next, simplify the denominator:
\( 1 - \frac{x}{x+1} = \frac{x+1 - x}{x+1} = \frac{1}{x+1} \).
Now, substitute these back into the \( \tan^{-1} \) expression:
\( L.H.S. = \tan^{-1} \frac{\frac{x^2+x+1}{x+1}}{\frac{1}{x+1}} \).
When dividing fractions, we multiply by the reciprocal of the denominator:
\( L.H.S. = \tan^{-1} \left(\frac{x^2+x+1}{x+1} \times \frac{x+1}{1}\right) \).
The \( (x+1) \) terms cancel out:
\( L.H.S. = \tan^{-1} (x^2+x+1) \).
This is equal to the Right Hand Side (R.H.S.). Hence proved.
In simple words: To prove this, we changed the \( \cot^{-1} \) part into an equivalent \( \tan^{-1} \) expression. Then, we used the formula for adding two inverse tangents. After carefully simplifying the fractions inside the formula, the entire left side turned into \( \tan^{-1} (x^2+x+1) \), which matched the right side of the equation.

🎯 Exam Tip: When a problem involves both \( \tan^{-1} \) and \( \cot^{-1} \), it's generally best to convert one to the other (e.g., \( \cot^{-1} A = \tan^{-1} \frac{1}{A} \)) to apply standard sum/difference formulas for inverse tangents more easily.

Question 16. If \( \alpha, \beta, \gamma \) be the roots of equation \( x^3 + px^2 + qx + p = 0 \), then prove that \( \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = n\pi \) except a special condition, when it does so on. Also find the special condition, when it does so on.
Answer: Given that \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 + px^2 + qx + p = 0 \). From Vieta's formulas, we know the sum and product of roots. The sum of the roots \( \alpha + \beta + \gamma \) is equal to \( -(\text{coefficient of } x^2 / \text{coefficient of } x^3) = -(p/1) = -p \). The sum of the product of roots taken two at a time, \( \alpha\beta + \beta\gamma + \gamma\alpha \), is equal to \( (\text{coefficient of } x / \text{coefficient of } x^3) = (q/1) = q \). Finally, the product of the roots \( \alpha\beta\gamma \) is equal to \( -(\text{constant term} / \text{coefficient of } x^3) = -(p/1) = -p \).
We need to prove that \( \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = n\pi \). The general formula for \( \tan^{-1}x + \tan^{-1}y + \tan^{-1}z \) is \( \tan^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}\right) \).
Substituting the values from the given equation's roots:
\( \alpha+\beta+\gamma = -p \)
\( \alpha\beta+\beta\gamma+\gamma\alpha = q \)
\( \alpha\beta\gamma = -p \)
Now, we substitute these into the tangent sum formula:
\( \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = \tan^{-1}\left(\frac{(\alpha+\beta+\gamma) - \alpha\beta\gamma}{1-(\alpha\beta+\beta\gamma+\gamma\alpha)}\right) \)
\( \implies \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = \tan^{-1}\left(\frac{(-p) - (-p)}{1-q}\right) \)
\( \implies \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = \tan^{-1}\left(\frac{-p+p}{1-q}\right) \)
\( \implies \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = \tan^{-1}\left(\frac{0}{1-q}\right) \)
\( \implies \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = \tan^{-1}(0) \)
We know that \( \tan^{-1}(0) = n\pi \) where \( n \) is an integer. Thus, \( \tan^{-1}\alpha + \tan^{-1}\beta + \tan^{-1}\gamma = n\pi \). This proof holds true as long as the denominator \( 1-q \neq 0 \).
The special condition when this identity might not hold, or "when it does so on", is when the denominator becomes zero, i.e., \( 1 - q = 0 \). This means \( q = 1 \). In this case, the expression becomes \( \tan^{-1}(0/0) \), which is undefined. If \( q=1 \), it implies that \( \alpha\beta + \beta\gamma + \gamma\alpha = 1 \). When \( q=1 \), the individual values of \( \tan^{-1}\alpha, \tan^{-1}\beta, \tan^{-1}\gamma \) might still be defined, but their sum cannot be simply found using this specific formula, and the relation \( = n\pi \) might not hold in all cases or may require a different form of the identity. The property of \( \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = n\pi \) is a fundamental result in inverse trigonometry.
In simple words: For the given equation, if we add up the inverse tangents of its roots, the result is always a multiple of \( \pi \). This works unless the value \( q \) in the equation is 1. If \( q \) is 1, the formula changes.

🎯 Exam Tip: Remember Vieta's formulas which link the coefficients of a polynomial to the sums and products of its roots. This is crucial for problems involving roots of equations and inverse trigonometric sums.

Question 17. Solve the following equations: (Q. 17 to 25)
\( \sec^{-1}\left(\frac{x}{a}\right) - \sec^{-1}\left(\frac{x}{b}\right) = \sec^{-1}b - \sec^{-1}a \)
Answer: We are given the equation:
\( \sec^{-1}\left(\frac{x}{a}\right) - \sec^{-1}\left(\frac{x}{b}\right) = \sec^{-1}b - \sec^{-1}a \)
First, we rearrange the terms to group similar inverse secant functions:
\( \sec^{-1}\left(\frac{x}{a}\right) + \sec^{-1}a = \sec^{-1}\left(\frac{x}{b}\right) + \sec^{-1}b \)
We know that \( \sec^{-1}y = \cos^{-1}\left(\frac{1}{y}\right) \). Applying this identity:
\( \cos^{-1}\left(\frac{a}{x}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{b}{x}\right) + \cos^{-1}\left(\frac{1}{b}\right) \)
Now, we use the formula \( \cos^{-1}A + \cos^{-1}B = \cos^{-1}(AB - \sqrt{1-A^2}\sqrt{1-B^2}) \).
Applying this to both sides:
\( \cos^{-1}\left[\frac{a}{x} \cdot \frac{1}{a} - \sqrt{1-\left(\frac{a}{x}\right)^2}\sqrt{1-\left(\frac{1}{a}\right)^2}\right] = \cos^{-1}\left[\frac{b}{x} \cdot \frac{1}{b} - \sqrt{1-\left(\frac{b}{x}\right)^2}\sqrt{1-\left(\frac{1}{b}\right)^2}\right] \)
\( \implies \cos^{-1}\left[\frac{1}{x} - \sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac{1}{a^2}}\right] = \cos^{-1}\left[\frac{1}{x} - \sqrt{1-\frac{b^2}{x^2}}\sqrt{1-\frac{1}{b^2}}\right] \)
Since \( \cos^{-1}A = \cos^{-1}B \implies A=B \), we can equate the arguments:
\( \frac{1}{x} - \sqrt{\frac{x^2-a^2}{x^2}}\sqrt{\frac{a^2-1}{a^2}} = \frac{1}{x} - \sqrt{\frac{x^2-b^2}{x^2}}\sqrt{\frac{b^2-1}{b^2}} \)
Cancel \( \frac{1}{x} \) from both sides and remove the negative sign:
\( \sqrt{\frac{x^2-a^2}{x^2}}\sqrt{\frac{a^2-1}{a^2}} = \sqrt{\frac{x^2-b^2}{x^2}}\sqrt{\frac{b^2-1}{b^2}} \)
Square both sides to remove the square roots:
\( \frac{(x^2-a^2)(a^2-1)}{x^2 a^2} = \frac{(x^2-b^2)(b^2-1)}{x^2 b^2} \)
Multiply both sides by \( x^2 \):
\( \frac{(x^2-a^2)(a^2-1)}{a^2} = \frac{(x^2-b^2)(b^2-1)}{b^2} \)
Cross-multiply:
\( b^2(x^2-a^2)(a^2-1) = a^2(x^2-b^2)(b^2-1) \)
\( b^2(x^2a^2 - x^2 - a^4 + a^2) = a^2(x^2b^2 - x^2 - b^4 + b^2) \)
\( x^2a^2b^2 - x^2b^2 - a^4b^2 + a^2b^2 = x^2a^2b^2 - x^2a^2 - a^2b^4 + a^2b^2 \)
Cancel \( x^2a^2b^2 \) and \( a^2b^2 \) from both sides:
\( -x^2b^2 - a^4b^2 = -x^2a^2 - a^2b^4 \)
Move terms with \( x^2 \) to one side and others to the other side:
\( x^2a^2 - x^2b^2 = a^4b^2 - a^2b^4 \)
Factor out \( x^2 \) on the left and \( a^2b^2 \) on the right:
\( x^2(a^2 - b^2) = a^2b^2(a^2 - b^2) \)
If \( a^2 - b^2 \neq 0 \), we can divide both sides by \( (a^2 - b^2) \):
\( x^2 = a^2b^2 \)
Take the square root of both sides:
\( x = \pm ab \)
If \( a^2 - b^2 = 0 \), i.e., \( a = \pm b \), then the original equation would simplify further or have different implications, but for the general case, \( x = \pm ab \) are the solutions. This problem involves simplifying inverse trigonometric expressions to find the value of x.
In simple words: We changed the inverse secant terms to inverse cosine terms and then used a formula to combine them. After simplifying the equation by removing square roots and common parts, we found that \( x \) can be either \( ab \) or \( -ab \).

🎯 Exam Tip: When solving inverse trigonometric equations, always try to convert all functions to the same type (e.g., all tan or all cos) and then use the sum/difference formulas to simplify.

Question 18. \( \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) + \tan^{-1}\left(\frac{2x}{x^2-1}\right) = \frac{2\pi}{3} \)
Answer: We are given the equation:
\( \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) + \tan^{-1}\left(\frac{2x}{x^2-1}\right) = \frac{2\pi}{3} \)
We know the identities: \( 2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) and \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \).
Let's rewrite the given terms using these identities. Note that \( \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) = \cos^{-1}\left(-\frac{1-x^2}{1+x^2}\right) \).
Using \( \cos^{-1}(-A) = \pi - \cos^{-1}A \):
\( \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) = \pi - \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \pi - 2\tan^{-1}x \)
For the second term: \( \tan^{-1}\left(\frac{2x}{x^2-1}\right) = \tan^{-1}\left(-\frac{2x}{1-x^2}\right) \).
Using \( \tan^{-1}(-A) = -\tan^{-1}A \):
\( \tan^{-1}\left(\frac{2x}{x^2-1}\right) = -\tan^{-1}\left(\frac{2x}{1-x^2}\right) = -2\tan^{-1}x \)
Substitute these back into the original equation:
\( (\pi - 2\tan^{-1}x) + (-2\tan^{-1}x) = \frac{2\pi}{3} \)
\( \implies \pi - 4\tan^{-1}x = \frac{2\pi}{3} \)
\( \implies \pi - \frac{2\pi}{3} = 4\tan^{-1}x \)
\( \implies \frac{3\pi - 2\pi}{3} = 4\tan^{-1}x \)
\( \implies \frac{\pi}{3} = 4\tan^{-1}x \)
\( \implies \tan^{-1}x = \frac{\pi}{12} \)
To find \( x \), take the tangent of both sides:
\( x = \tan\left(\frac{\pi}{12}\right) \)
We know that \( \tan(\frac{\pi}{12}) = \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \).
\( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
So, \( x = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \).
Rationalize the denominator:
\( x = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3 + 1 - 2\sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \).
Thus, the solution is \( x = 2 - \sqrt{3} \). The use of standard inverse trigonometric identities simplifies the problem significantly.
In simple words: We changed the given equation into a simpler form using special math rules for inverse tangent and cosine. After moving terms around, we found that the inverse tangent of \( x \) is \( \frac{\pi}{12} \). By taking the tangent, we got \( x = 2 - \sqrt{3} \).

🎯 Exam Tip: Recognizing standard inverse trigonometric identities like \( 2\tan^{-1}x \) in different forms is key to quickly solving such problems. Also, know the values of trigonometric functions for common angles like \( \pi/12 \).

Question 19. \( \tan^{-1}\left(\frac{1}{1+2x}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \tan^{-1}\left(\frac{2}{x^2}\right) \)
Answer: We are given the equation:
\( \tan^{-1}\left(\frac{1}{1+2x}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \tan^{-1}\left(\frac{2}{x^2}\right) \)
We use the formula \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \) on the left-hand side.
Let \( A = \frac{1}{1+2x} \) and \( B = \frac{1}{4x+1} \).
L.H.S. \( = \tan^{-1}\left(\frac{\frac{1}{1+2x} + \frac{1}{4x+1}}{1 - \frac{1}{1+2x} \cdot \frac{1}{4x+1}}\right) \)
\( = \tan^{-1}\left(\frac{\frac{(4x+1)+(1+2x)}{(1+2x)(4x+1)}}{\frac{(1+2x)(4x+1)-1}{(1+2x)(4x+1)}}\right) \)
\( = \tan^{-1}\left(\frac{4x+1+1+2x}{(1+2x)(4x+1)-1}\right) \)
\( = \tan^{-1}\left(\frac{6x+2}{4x+1+8x^2+2x-1}\right) \)
\( = \tan^{-1}\left(\frac{6x+2}{8x^2+6x}\right) \)
\( = \tan^{-1}\left(\frac{2(3x+1)}{2x(4x+3)}\right) \)
\( = \tan^{-1}\left(\frac{3x+1}{x(4x+3)}\right) \)
Now, equate this to the right-hand side of the original equation:
\( \tan^{-1}\left(\frac{3x+1}{x(4x+3)}\right) = \tan^{-1}\left(\frac{2}{x^2}\right) \)
Since the inverse tangent functions are equal, their arguments must be equal:
\( \frac{3x+1}{x(4x+3)} = \frac{2}{x^2} \)
We can cancel one \( x \) from the denominator on both sides, assuming \( x \neq 0 \):
\( \frac{3x+1}{4x+3} = \frac{2}{x} \)
Cross-multiply:
\( x(3x+1) = 2(4x+3) \)
\( 3x^2+x = 8x+6 \)
Rearrange into a quadratic equation:
\( 3x^2 + x - 8x - 6 = 0 \)
\( 3x^2 - 7x - 6 = 0 \)
We can solve this quadratic equation by factoring or using the quadratic formula.
To factor, look for two numbers that multiply to \( 3 \times (-6) = -18 \) and add to \( -7 \). These numbers are \( -9 \) and \( 2 \).
\( 3x^2 - 9x + 2x - 6 = 0 \)
\( 3x(x-3) + 2(x-3) = 0 \)
\( (x-3)(3x+2) = 0 \)
This gives two possible solutions:
\( x-3 = 0 \implies x = 3 \)
\( 3x+2 = 0 \implies x = -\frac{2}{3} \)
We must check these solutions in the original equation to ensure that the arguments of \( \tan^{-1} \) are valid. Both values yield defined arguments. The process involves algebraic manipulation of inverse trigonometric functions.
In simple words: We combined the two inverse tangent terms on the left side using a special math rule. Then, we set this equal to the right side and simplified the equation. This gave us a quadratic equation, which we solved to find two possible values for \( x \): \( 3 \) and \( -\frac{2}{3} \).

🎯 Exam Tip: Always simplify fractions and common factors before cross-multiplication. Remember to check for any restrictions on the domain of the inverse trigonometric functions, especially when squaring or simplifying expressions with variables in the denominator.

Question 20. \( \tan^{-1}\left(\frac{x+7}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \pi - \tan^{-1}7 \)
Answer: We are given the equation:
\( \tan^{-1}\left(\frac{x+7}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \pi - \tan^{-1}7 \)
First, move \( \tan^{-1}7 \) to the left-hand side:
\( \tan^{-1}\left(\frac{x+7}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) + \tan^{-1}7 = \pi \)
We use the formula \( \tan^{-1}A + \tan^{-1}B + \tan^{-1}C = \tan^{-1}\left(\frac{A+B+C-ABC}{1-(AB+BC+CA)}\right) \).
Let \( A = \frac{x+7}{x-1} \), \( B = \frac{x-1}{x} \), and \( C = 7 \).
First, let's calculate \( A+B \):
\( \frac{x+7}{x-1} + \frac{x-1}{x} = \frac{x(x+7) + (x-1)^2}{x(x-1)} = \frac{x^2+7x + x^2-2x+1}{x(x-1)} = \frac{2x^2+5x+1}{x^2-x} \)
Now, calculate \( AB \):
\( \left(\frac{x+7}{x-1}\right) \left(\frac{x-1}{x}\right) = \frac{x+7}{x} \)
Using the formula for two terms first: \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \)
\( \tan^{-1}\left(\frac{x+7}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \tan^{-1}\left(\frac{\frac{2x^2+5x+1}{x^2-x}}{1-\frac{x+7}{x}}\right) \)
\( = \tan^{-1}\left(\frac{\frac{2x^2+5x+1}{x^2-x}}{\frac{x-(x+7)}{x}}\right) = \tan^{-1}\left(\frac{\frac{2x^2+5x+1}{x^2-x}}{\frac{-7}{x}}\right) \)
\( = \tan^{-1}\left(\frac{2x^2+5x+1}{x(x-1)} \cdot \frac{x}{-7}\right) = \tan^{-1}\left(\frac{2x^2+5x+1}{-7(x-1)}\right) = \tan^{-1}\left(\frac{2x^2+5x+1}{-7x+7}\right) \)
So, the equation becomes:
\( \tan^{-1}\left(\frac{2x^2+5x+1}{-7x+7}\right) + \tan^{-1}7 = \pi \)
Apply the \( \tan^{-1}A + \tan^{-1}B \) formula again:
\( \tan^{-1}\left(\frac{\frac{2x^2+5x+1}{-7x+7} + 7}{1 - \left(\frac{2x^2+5x+1}{-7x+7}\right) \cdot 7}\right) = \pi \)
Taking \( \tan \) on both sides:
\( \frac{\frac{2x^2+5x+1}{-7x+7} + 7}{1 - \frac{7(2x^2+5x+1)}{-7x+7}} = \tan(\pi) \)
We know \( \tan(\pi) = 0 \). So the numerator must be zero:
\( \frac{2x^2+5x+1}{-7x+7} + 7 = 0 \)
\( \frac{2x^2+5x+1 + 7(-7x+7)}{-7x+7} = 0 \)
\( 2x^2+5x+1 - 49x+49 = 0 \)
\( 2x^2 - 44x + 50 = 0 \)
Divide by 2:
\( x^2 - 22x + 25 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \):
\( x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4(1)(25)}}{2(1)} \)
\( x = \frac{22 \pm \sqrt{484 - 100}}{2} \)
\( x = \frac{22 \pm \sqrt{384}}{2} \)
We know \( \sqrt{384} = \sqrt{64 \times 6} = 8\sqrt{6} \).
\( x = \frac{22 \pm 8\sqrt{6}}{2} \)
\( x = 11 \pm 4\sqrt{6} \)
These solutions represent the values of x for which the inverse trigonometric identity holds.
In simple words: We moved all inverse tangent terms to one side, setting the sum equal to \( \pi \). Then, we used the formula for adding inverse tangents. Since \( \tan(\pi) \) is zero, we set the numerator of the combined term to zero. This gave us a quadratic equation, which we solved to get \( x = 11 \pm 4\sqrt{6} \).

🎯 Exam Tip: When the sum of inverse tangents equals \( \pi \), it implies the numerator of the combined tangent formula must be zero (assuming the denominator is non-zero). Remember to simplify square roots by finding perfect square factors.

Question 21. \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \cot^{-1}x = \frac{\pi}{4} \)
Answer: We are given the equation:
\( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \cot^{-1}x = \frac{\pi}{4} \)
To solve this, it's often easier to convert all inverse trigonometric functions to a single type, usually \( \tan^{-1} \).
Let \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = \theta \). Then \( \sin\theta = \frac{1}{\sqrt{5}} \).
We can form a right-angled triangle where the opposite side is 1 and the hypotenuse is \( \sqrt{5} \).
Using Pythagoras theorem, the adjacent side is \( \sqrt{(\sqrt{5})^2 - 1^2} = \sqrt{5-1} = \sqrt{4} = 2 \).
So, \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{2} \). Therefore, \( \theta = \tan^{-1}\left(\frac{1}{2}\right) \).
Also, we know that \( \cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right) \) for \( x > 0 \). (Assuming x is positive, if x were negative or zero, we would need to consider cases, but usually, these problems assume positive principal values).
Substitute these back into the original equation:
\( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \)
Use the formula \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \):
\( \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{x}}{1 - \frac{1}{2} \cdot \frac{1}{x}}\right) = \frac{\pi}{4} \)
\( \implies \tan^{-1}\left(\frac{\frac{x+2}{2x}}{\frac{2x-1}{2x}}\right) = \frac{\pi}{4} \)
\( \implies \tan^{-1}\left(\frac{x+2}{2x-1}\right) = \frac{\pi}{4} \)
Take the tangent of both sides:
\( \frac{x+2}{2x-1} = \tan\left(\frac{\pi}{4}\right) \)
We know \( \tan\left(\frac{\pi}{4}\right) = 1 \).
\( \frac{x+2}{2x-1} = 1 \)
Cross-multiply:
\( x+2 = 2x-1 \)
Move \( x \) terms to one side and constants to the other:
\( 2+1 = 2x-x \)
\( 3 = x \)
So, the solution to the equation is \( x=3 \). The ability to convert between different inverse trigonometric functions is essential here.
In simple words: First, we changed \( \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \) into its equivalent \( \tan^{-1} \) form, which is \( \tan^{-1}\left(\frac{1}{2}\right) \). We also changed \( \cot^{-1}x \) to \( \tan^{-1}\left(\frac{1}{x}\right) \). Then, we added these two \( \tan^{-1} \) terms using a formula and set the result equal to \( \frac{\pi}{4} \). Since \( \tan(\frac{\pi}{4}) = 1 \), we solved the simple equation to find \( x=3 \).

🎯 Exam Tip: Always convert all inverse trigonometric terms to a single function (preferably \( \tan^{-1} \)) before applying sum/difference formulas. Remember the basic trigonometric ratios for common angles like \( \pi/4 \).

Question 22. \( 3\tan^{-1}\left(\frac{1}{2+\sqrt{3}}\right) - \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{3}\right) \)
Answer: We are given the equation:
\( 3\tan^{-1}\left(\frac{1}{2+\sqrt{3}}\right) - \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{3}\right) \)
First, simplify the argument of the first inverse tangent term. We know that \( 2+\sqrt{3} \) is related to \( \tan\left(\frac{\pi}{12}\right) \). Specifically, \( \frac{1}{2+\sqrt{3}} = \frac{1}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3} \).
We know that \( \tan\left(\frac{\pi}{12}\right) = 2-\sqrt{3} \).
So, \( \tan^{-1}\left(\frac{1}{2+\sqrt{3}}\right) = \tan^{-1}(2-\sqrt{3}) = \frac{\pi}{12} \).
Substitute this back into the equation:
\( 3\left(\frac{\pi}{12}\right) - \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{3}\right) \)
\( \implies \frac{\pi}{4} - \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{3}\right) \)
Rearrange the terms:
\( \frac{\pi}{4} = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{x}\right) \)
Use the formula \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \) on the right-hand side:
\( \frac{\pi}{4} = \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{x}}{1 - \frac{1}{3} \cdot \frac{1}{x}}\right) \)
\( \implies \frac{\pi}{4} = \tan^{-1}\left(\frac{\frac{x+3}{3x}}{\frac{3x-1}{3x}}\right) \)
\( \implies \frac{\pi}{4} = \tan^{-1}\left(\frac{x+3}{3x-1}\right) \)
Take the tangent of both sides:
\( \tan\left(\frac{\pi}{4}\right) = \frac{x+3}{3x-1} \)
We know \( \tan\left(\frac{\pi}{4}\right) = 1 \).
\( 1 = \frac{x+3}{3x-1} \)
Cross-multiply:
\( 3x-1 = x+3 \)
Move \( x \) terms to one side and constants to the other:
\( 3x-x = 3+1 \)
\( 2x = 4 \)
\( x = 2 \)
So, the solution to the equation is \( x=2 \). Recognizing the special value of the tangent simplifies the problem greatly.
In simple words: First, we simplified \( \tan^{-1}\left(\frac{1}{2+\sqrt{3}}\right) \) to \( \frac{\pi}{12} \) because \( \frac{1}{2+\sqrt{3}} \) is the same as \( 2-\sqrt{3} \), which is \( \tan(\frac{\pi}{12}) \). After this, the equation became \( \frac{\pi}{4} - \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{3}\right) \). We rearranged it to solve for \( \tan^{-1}\left(\frac{1}{x}\right) \), then applied the formula for adding inverse tangents. Finally, by taking the tangent of both sides, we found \( x=2 \).

🎯 Exam Tip: Familiarize yourself with special trigonometric values like \( \tan(\pi/12) \) and \( \tan(\pi/8) \). Rationalizing denominators with surds is often the first step in simplifying such inverse trigonometric arguments.

Question 23. \( \sin\left[2\cos^{-1}\left\{\cot(2\tan^{-1}x)\right\}\right] = 0 \)
Answer: We are given the equation:
\( \sin\left[2\cos^{-1}\left\{\cot(2\tan^{-1}x)\right\}\right] = 0 \)
For \( \sin \theta = 0 \), we know that \( \theta = n\pi \) for some integer \( n \).
So, \( 2\cos^{-1}\left\{\cot(2\tan^{-1}x)\right\} = n\pi \).
Let's simplify from the innermost function, \( 2\tan^{-1}x \). We use the identity \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \).
So, \( \cot(2\tan^{-1}x) = \cot\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) \).
We know that \( \cot(\tan^{-1}A) = \frac{1}{A} \).
So, \( \cot\left(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right) = \frac{1-x^2}{2x} \).
Substitute this back into the equation:
\( 2\cos^{-1}\left(\frac{1-x^2}{2x}\right) = n\pi \)
\( \implies \cos^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{n\pi}{2} \)
For \( \cos^{-1}y \) to be defined, \( -1 \le y \le 1 \). Also, the range of \( \cos^{-1} \) is \( [0, \pi] \).
So, \( \frac{n\pi}{2} \) must be in the range \( [0, \pi] \). This means \( n \) can be \( 0, 1, 2 \).
Case 1: \( n=0 \)
\( \cos^{-1}\left(\frac{1-x^2}{2x}\right) = 0 \)
\( \frac{1-x^2}{2x} = \cos(0) = 1 \)
\( 1-x^2 = 2x \)
\( x^2 + 2x - 1 = 0 \)
Using the quadratic formula:
\( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4+4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \).
Case 2: \( n=1 \)
\( \cos^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2} \)
\( \frac{1-x^2}{2x} = \cos\left(\frac{\pi}{2}\right) = 0 \)
\( 1-x^2 = 0 \)
\( x^2 = 1 \implies x = \pm 1 \).
Case 3: \( n=2 \)
\( \cos^{-1}\left(\frac{1-x^2}{2x}\right) = \pi \)
\( \frac{1-x^2}{2x} = \cos(\pi) = -1 \)
\( 1-x^2 = -2x \)
\( x^2 - 2x - 1 = 0 \)
Using the quadratic formula:
\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \).
Combining all possible solutions, we have \( x = \pm 1, \pm(1-\sqrt{2}), \pm(1+\sqrt{2}) \). We must also consider the domain restriction for \( \tan^{-1}x \), which is all real numbers. However, the identity \( \cot(2\tan^{-1}x) = \frac{1-x^2}{2x} \) is valid when \( x \neq 0, \pm 1 \). If \( x=1 \) or \( x=-1 \), \( 1-x^2=0 \). In this case, \( \cot(2\tan^{-1}(\pm 1)) = \cot(\pm \frac{\pi}{2}) = 0 \). So \( 2\cos^{-1}(0) = 2(\frac{\pi}{2}) = \pi \), which matches \( n=2 \). So \( x=\pm 1 \) are valid solutions. The solution involves carefully applying inverse trigonometric identities and solving resulting algebraic equations.
In simple words: Since \( \sin(\text{angle}) = 0 \), the angle inside must be a multiple of \( \pi \). We simplified the complex expression inside the \( \sin \) function, starting from the innermost part. We used formulas to change \( 2\tan^{-1}x \) and \( \cot(\tan^{-1}A) \). This led to \( \cos^{-1}(\frac{1-x^2}{2x}) \) being a multiple of \( \frac{\pi}{2} \). By considering the possible values for this multiple (\( 0, \frac{\pi}{2}, \pi \)), we solved three different equations for \( x \), getting \( \pm 1 \), \( -1 \pm \sqrt{2} \) and \( 1 \pm \sqrt{2} \).

🎯 Exam Tip: Break down complex inverse trigonometric equations by simplifying from the inside out. Pay close attention to the range of inverse trigonometric functions and the values of trigonometric functions at multiples of \( \pi/2 \).

Question 24. \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) + \tan^{-1}\left(\frac{1}{6}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \)
Answer: We are given the equation:
\( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) + \tan^{-1}\left(\frac{1}{6}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \)
We will combine the inverse tangent terms on the left-hand side. It's usually easier to combine them in pairs.
First, combine \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \):
\( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9}}\right) = \tan^{-1}\left(\frac{\frac{9+8}{36}}{1 - \frac{2}{36}}\right) = \tan^{-1}\left(\frac{\frac{17}{36}}{\frac{34}{36}}\right) = \tan^{-1}\left(\frac{17}{34}\right) = \tan^{-1}\left(\frac{1}{2}\right) \).
Now, the equation becomes:
\( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{6}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \)
Combine \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{6}\right) \):
\( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{6}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{6}}{1 - \frac{1}{2} \cdot \frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{3+1}{6}}{1 - \frac{1}{12}}\right) = \tan^{-1}\left(\frac{\frac{4}{6}}{\frac{11}{12}}\right) = \tan^{-1}\left(\frac{2}{3} \cdot \frac{12}{11}\right) = \tan^{-1}\left(\frac{8}{11}\right) \).
The equation simplifies to:
\( \tan^{-1}\left(\frac{8}{11}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \)
Now, combine these two terms:
\( \tan^{-1}\left(\frac{\frac{8}{11} + \frac{1}{x}}{1 - \frac{8}{11} \cdot \frac{1}{x}}\right) = \frac{\pi}{4} \)
\( \implies \tan^{-1}\left(\frac{\frac{8x+11}{11x}}{\frac{11x-8}{11x}}\right) = \frac{\pi}{4} \)
\( \implies \tan^{-1}\left(\frac{8x+11}{11x-8}\right) = \frac{\pi}{4} \)
Take the tangent of both sides:
\( \frac{8x+11}{11x-8} = \tan\left(\frac{\pi}{4}\right) \)
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \):
\( \frac{8x+11}{11x-8} = 1 \)
Cross-multiply:
\( 8x+11 = 11x-8 \)
Move \( x \) terms to one side and constants to the other:
\( 11+8 = 11x-8x \)
\( 19 = 3x \)
\( x = \frac{19}{3} \)
Thus, the solution to the equation is \( x = \frac{19}{3} \). Combining terms systematically simplifies the problem.
In simple words: We added the inverse tangent terms on the left side in pairs using a specific formula. First, \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \) simplified to \( \tan^{-1}\left(\frac{1}{2}\right) \). Then, we added this to \( \tan^{-1}\left(\frac{1}{6}\right) \), which simplified to \( \tan^{-1}\left(\frac{8}{11}\right) \). Finally, we added \( \tan^{-1}\left(\frac{8}{11}\right) \) to \( \tan^{-1}\left(\frac{1}{x}\right) \) and set it equal to \( \frac{\pi}{4} \). Since \( \tan(\frac{\pi}{4})=1 \), we solved the resulting algebraic equation to find \( x = \frac{19}{3} \).

🎯 Exam Tip: When dealing with multiple inverse tangent terms, combine them in a systematic way (e.g., two by two) to avoid errors. Always simplify fractions completely at each step.

Question 25. If \( \sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} \) and \( \cos^{-1}x - \cos^{-1}y = \frac{\pi}{3} \), then find the values of \( x \) and \( y \).
Answer: We are given two equations:
1. \( \sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} \)
2. \( \cos^{-1}x - \cos^{-1}y = \frac{\pi}{3} \)
We know the identity \( \sin^{-1}A + \cos^{-1}A = \frac{\pi}{2} \). So, \( \cos^{-1}A = \frac{\pi}{2} - \sin^{-1}A \).
Let's substitute this into the second equation:
\( \left(\frac{\pi}{2} - \sin^{-1}x\right) - \left(\frac{\pi}{2} - \sin^{-1}y\right) = \frac{\pi}{3} \)
\( \implies \frac{\pi}{2} - \sin^{-1}x - \frac{\pi}{2} + \sin^{-1}y = \frac{\pi}{3} \)
\( \implies -\sin^{-1}x + \sin^{-1}y = \frac{\pi}{3} \)
Now we have a system of two linear equations with \( \sin^{-1}x \) and \( \sin^{-1}y \):
(A) \( \sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} \)
(B) \( -\sin^{-1}x + \sin^{-1}y = \frac{\pi}{3} \)
Add equation (A) and (B):
\( (\sin^{-1}x + \sin^{-1}y) + (-\sin^{-1}x + \sin^{-1}y) = \frac{2\pi}{3} + \frac{\pi}{3} \)
\( \implies 2\sin^{-1}y = \pi \)
\( \implies \sin^{-1}y = \frac{\pi}{2} \)
Now, find \( y \):
\( y = \sin\left(\frac{\pi}{2}\right) = 1 \)
Substitute \( \sin^{-1}y = \frac{\pi}{2} \) back into equation (A):
\( \sin^{-1}x + \frac{\pi}{2} = \frac{2\pi}{3} \)
\( \implies \sin^{-1}x = \frac{2\pi}{3} - \frac{\pi}{2} \)
To subtract, find a common denominator (6):
\( \sin^{-1}x = \frac{4\pi}{6} - \frac{3\pi}{6} = \frac{\pi}{6} \)
Now, find \( x \):
\( x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)
So, the values are \( x = \frac{1}{2} \) and \( y = 1 \). This problem showcases how to solve simultaneous equations involving inverse trigonometric functions.
In simple words: We had two equations with inverse sine and inverse cosine. We used the rule that \( \cos^{-1}A = \frac{\pi}{2} - \sin^{-1}A \) to change the second equation so it only had inverse sines. Then, we had two simple equations with \( \sin^{-1}x \) and \( \sin^{-1}y \). By adding and subtracting these equations, we found \( \sin^{-1}y = \frac{\pi}{2} \) and \( \sin^{-1}x = \frac{\pi}{6} \). From these, we got \( y=1 \) and \( x=\frac{1}{2} \).

🎯 Exam Tip: When solving simultaneous inverse trigonometric equations, convert all functions to a single type (e.g., all \( \sin^{-1} \) or all \( \tan^{-1} \)) to simplify the system into linear equations. Remember the fundamental identity \( \sin^{-1}A + \cos^{-1}A = \pi/2 \).