CBSE Class 12 Mathematics Inverse Trigonometric Functions Assignment Set 01

CBSE Class 12 Mathematics Inverse Trigonometric Functions Assignment Set A. Chapter wise assignments are being given by teachers to students to make them understand the chapter concepts. Its extremely critical for all CBSE students to practice all assignments which will help them in gaining better marks in examinations. All assignments available for free download on the website are developed by the best teachers having many years of teaching experience in CBSE schools all over the country. Students, teachers and parents are advised to download the CBSE study material.

Inverse Trigonometric Functions

Important Points

• The expressions \( \sin^{-1} x \), \( \cos^{-1} x \), etc., represent angles.
• If \( \sin \theta = x \) and \( \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), then we can write \( \theta = \sin^{-1} x \). Similar conditions apply to the other inverse trigonometric functions.

Domain and Range of Inverse Trigonometric Functions

Basic Properties of Inverse Trigonometric Functions

• \( \sin^{-1} (\sin x) = x \) for all \( x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)
• \( \cos^{-1} (\cos x) = x \) for all \( x \in [0, \pi] \)
• \( \sin (\sin^{-1} x) = x \) for all \( x \in [-1, 1] \)
• \( \cos (\cos^{-1} x) = x \) for all \( x \in [-1, 1] \)

Reciprocal Relations

• \( \sin^{-1} x = \csc^{-1} \left( \frac{1}{x} \right) \) for all \( x \in [-1, 1] \setminus \{0\} \)
• \( \tan^{-1} x = \cot^{-1} \left( \frac{1}{x} \right) \) for all \( x > 0 \)
• \( \sec^{-1} x = \cos^{-1} \left( \frac{1}{x} \right) \) for all \( |x| \ge 1 \)

Negatives of Arguments

• \( \sin^{-1}(-x) = -\sin^{-1} x \) for all \( x \in [-1, 1] \)
• \( \tan^{-1}(-x) = -\tan^{-1} x \) for all \( x \in \mathbb{R} \)
• \( \csc^{-1}(-x) = -\csc^{-1} x \) for all \( |x| \ge 1 \)
• \( \cos^{-1}(-x) = \pi - \cos^{-1} x \) for all \( x \in [-1, 1] \)
• \( \cot^{-1}(-x) = \pi - \cot^{-1} x \) for all \( x \in \mathbb{R} \)
• \( \sec^{-1}(-x) = \pi - \sec^{-1} x \) for all \( |x| \ge 1 \)

Complementary Angles Sum Identities

• \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) for all \( x \in [-1, 1] \)
• \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \) for all \( x \in \mathbb{R} \)
• \( \sec^{-1} x + \csc^{-1} x = \frac{\pi}{2} \) for all \( |x| \ge 1 \)

Sum and Difference of Tan Inverse

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

Double Angle Identities for Tan Inverse

• \( 2\tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right) \) when \( |x| < 1 \)
• \( 2\tan^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \) when \( |x| \le 1 \)
• \( 2\tan^{-1} x = \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) \) when \( x \ge 0 \)

Very Short Answer Type Questions (1 Mark)

Question 1. Write the principal value of:
(i) \( \sin^{-1}\left( -\frac{\sqrt{3}}{2} \right) \)
(ii) \( \cos^{-1}\left( \frac{\sqrt{3}}{2} \right) \)
(iii) \( \tan^{-1}\left( -\frac{1}{\sqrt{3}} \right) \)
(iv) \( \csc^{-1}(-2) \)
(v) \( \cot^{-1}\left( \frac{1}{\sqrt{3}} \right) \)
(vi) \( \sec^{-1}(-2) \)
(vii) \( \sin^{-1}\left( -\frac{\sqrt{3}}{2} \right) + \cos^{-1}\left( -\frac{1}{2} \right) + \tan^{-1}\left( -\frac{1}{\sqrt{3}} \right) \)
Answer:
(i) Let \( \theta = \sin^{-1}\left( -\frac{\sqrt{3}}{2} \right) \). Since the principal value branch of \( \sin^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we get \( \theta = -\frac{\pi}{3} \).
(ii) Let \( \theta = \cos^{-1}\left( \frac{\sqrt{3}}{2} \right) \). Since the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \), we get \( \theta = \frac{\pi}{6} \).
(iii) Let \( \theta = \tan^{-1}\left( -\frac{1}{\sqrt{3}} \right) \). Since the principal value branch of \( \tan^{-1} \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we get \( \theta = -\frac{\pi}{6} \).
(iv) Let \( \theta = \csc^{-1}(-2) \). Since the principal value branch of \( \csc^{-1} \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\} \), we get \( \theta = -\frac{\pi}{6} \).
(v) Let \( \theta = \cot^{-1}\left( \frac{1}{\sqrt{3}} \right) \). Since the principal value branch of \( \cot^{-1} \) is \( (0, \pi) \), we get \( \theta = \frac{\pi}{3} \).
(vi) Let \( \theta = \sec^{-1}(-2) \). Since the principal value branch of \( \sec^{-1} \) is \( [0, \pi] \setminus \left\{\frac{\pi}{2}\right\} \), we get \( \theta = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).
(vii) We evaluate each term using its principal value branch:
\( \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \)
\( \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)
\( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \)
Adding these values:
\( -\frac{\pi}{3} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \).
In simple words: To find the principal values, look for the angle within the standard range of each function that gives the target number.

Exam Tip: Memorise the principal value branch (ranges) of all six inverse trigonometric functions - this is crucial for getting full marks on 1-mark questions.

Question 2. What is the value of the following functions (using principal value):
(i) \( \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) - \sec^{-1}\left( \frac{2}{\sqrt{3}} \right) \)
(ii) \( \sin^{-1}\left( -\frac{1}{2} \right) - \cos^{-1}\left( \frac{\sqrt{3}}{2} \right) \)
(iii) \( \tan^{-1}(1) - \cot^{-1}(-1) \)
(iv) \( \csc^{-1}(\sqrt{2}) + \sec^{-1}(\sqrt{2}) \)
(v) \( \tan^{-1}(1) + \cot^{-1}(1) + \sin^{-1}(1) \)
(vi) \( \sin^{-1}\left( \sin \frac{4\pi}{5} \right) \)
(vii) \( \tan^{-1}\left( \tan \frac{5\pi}{6} \right) \)
(viii) \( \csc^{-1}\left( \csc \frac{3\pi}{4} \right) \)
Answer:
(i) Since \( \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \) and \( \sec^{-1}\left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6} \), we have:
\( \frac{\pi}{6} - \frac{\pi}{6} = 0 \).
(ii) Since \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \) and \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \), we have:
\( -\frac{\pi}{6} - \frac{\pi}{6} = -\frac{\pi}{3} \).
(iii) Since \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( \cot^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \), we have:
\( \frac{\pi}{4} - \frac{3\pi}{4} = -\frac{2\pi}{4} = -\frac{\pi}{2} \).
(iv) Using the identity \( \csc^{-1} x + \sec^{-1} x = \frac{\pi}{2} \), we immediately get:
\( \csc^{-1}(\sqrt{2}) + \sec^{-1}(\sqrt{2}) = \frac{\pi}{2} \).
(v) Using the identity \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \) and knowing \( \sin^{-1}(1) = \frac{\pi}{2} \), we have:
\( \left(\tan^{-1}(1) + \cot^{-1}(1)\right) + \sin^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \).
(vi) Since \( \frac{4\pi}{5} \) is outside the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we rewrite:
\( \sin\left(\frac{4\pi}{5}\right) = \sin\left(\pi - \frac{\pi}{5}\right) = \sin\left(\frac{\pi}{5}\right) \).
Thus, \( \sin^{-1}\left( \sin \frac{4\pi}{5} \right) = \sin^{-1}\left(\sin\frac{\pi}{5}\right) = \frac{\pi}{5} \) (as \( \frac{\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)).
(vii) Since \( \frac{5\pi}{6} \) is outside the range \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we rewrite:
\( \tan\left(\frac{5\pi}{6}\right) = \tan\left(\pi - \frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = \tan\left(-\frac{\pi}{6}\right) \).
Thus, \( \tan^{-1}\left( \tan \frac{5\pi}{6} \right) = \tan^{-1}\left(\tan\left(-\frac{\pi}{6}\right)\right) = -\frac{\pi}{6} \) (as \( -\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)).
(viii) Since \( \frac{3\pi}{4} \) is outside the range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\} \), we rewrite:
\( \csc\left(\frac{3\pi}{4}\right) = \csc\left(\pi - \frac{\pi}{4}\right) = \csc\left(\frac{\pi}{4}\right) \).
Thus, \( \csc^{-1}\left( \csc \frac{3\pi}{4} \right) = \csc^{-1}\left(\csc\frac{\pi}{4}\right) = \frac{\pi}{4} \) (as \( \frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\} \)).
In simple words: When the given angle is outside the standard range of the inverse function, use trigonometric identities like \( \sin(\pi - \theta) = \sin\theta \) to find an equivalent angle within the range.

Exam Tip: Be very careful when simplifying expression like \( f^{-1}(f(\theta)) \). Always verify if \( \theta \) lies in the principal value branch before direct cancellation.

Short Answer Type Questions (4 Marks)

Question 3. Show that \( \tan^{-1}\left( \frac{\sqrt{1 + \cos x} + \sqrt{1 - \cos x}}{\sqrt{1 + \cos x} - \sqrt{1 - \cos x}} \right) = \frac{\pi}{4} + \frac{x}{2} \), \( x \in [0, \pi] \)
Answer:
Let \( \text{LHS} = \tan^{-1}\left( \frac{\sqrt{1 + \cos x} + \sqrt{1 - \cos x}}{\sqrt{1 + \cos x} - \sqrt{1 - \cos x}} \right) \).
Using half-angle trigonometric formulas, we know:
\( 1 + \cos x = 2\cos^2\left(\frac{x}{2}\right) \)
\( 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \)
Since \( x \in [0, \pi] \), we have \( \frac{x}{2} \in \left[0, \frac{\pi}{2}\right] \), which lies in the first quadrant where both \( \cos\left(\frac{x}{2}\right) \) and \( \sin\left(\frac{x}{2}\right) \) are positive.
Thus:
\( \sqrt{1 + \cos x} = \sqrt{2}\cos\left(\frac{x}{2}\right) \)
\( \sqrt{1 - \cos x} = \sqrt{2}\sin\left(\frac{x}{2}\right) \)
Substitute these expressions back into the LHS:
\( \text{LHS} = \tan^{-1}\left( \frac{\sqrt{2}\cos(x/2) + \sqrt{2}\sin(x/2)}{\sqrt{2}\cos(x/2) - \sqrt{2}\sin(x/2)} \right) \)
\( = \tan^{-1}\left( \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} \right) \)
Divide both the numerator and the denominator by \( \cos(x/2) \):
\( = \tan^{-1}\left( \frac{1 + \tan(x/2)}{1 - \tan(x/2)} \right) \)
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we can write this as:
\( = \tan^{-1}\left( \frac{\tan(\pi/4) + \tan(x/2)}{1 - \tan(\pi/4)\tan(x/2)} \right) \)
Applying the sum formula \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\( = \tan^{-1}\left( \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \right) \)
Since \( x \in [0, \pi] \), \( \frac{\pi}{4} + \frac{x}{2} \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \). For the principal branch restriction of \( \tan^{-1} \), the identity holds, and we obtain:
\( = \frac{\pi}{4} + \frac{x}{2} = \text{RHS} \).
Hence proved.
In simple words: Replace the terms inside the square roots with half-angle identities to eliminate the roots, simplify the fraction to a tangent formula, and cancel the inverse tangent.

Exam Tip: Clearly write down the half-angle identities you used in a sidebar or bracket - examiners award specific marks for stating the auxiliary formulas.

Question 4. Prove \( \tan^{-1}\left( \frac{\cos x}{1 - \sin x} \right) - \cot^{-1}\left( \sqrt{\frac{1 + \cos x}{1 - \cos x}} \right) = \frac{\pi}{4} \), \( x \in (0, \pi/2) \).
Answer:
Let us evaluate the two terms separately.
First term: \( T_1 = \tan^{-1}\left( \frac{\cos x}{1 - \sin x} \right) \)
We can write:
\( \cos x = \cos^2(x/2) - \sin^2(x/2) = \left(\cos(x/2) - \sin(x/2)\right)\left(\cos(x/2) + \sin(x/2)\right) \)
And:
\( 1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = \left(\cos(x/2) - \sin(x/2)\right)^2 \)
Substitute these into the fraction:
\( \frac{\cos x}{1 - \sin x} = \frac{\left(\cos(x/2) - \sin(x/2)\right)\left(\cos(x/2) + \sin(x/2)\right)}{\left(\cos(x/2) - \sin(x/2)\right)^2} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} \)
Divide the numerator and denominator by \( \cos(x/2) \):
\( = \frac{1 + \tan(x/2)}{1 - \tan(x/2)} = \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \)
Since \( x \in (0, \pi/2) \), we have \( \frac{\pi}{4} + \frac{x}{2} \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \), which lies within the principal value branch of \( \tan^{-1} \).
Thus, \( T_1 = \tan^{-1}\left( \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \right) = \frac{\pi}{4} + \frac{x}{2} \).
Second term: \( T_2 = \cot^{-1}\left( \sqrt{\frac{1 + \cos x}{1 - \cos x}} \right) \)
Using half-angle identities:
\( 1 + \cos x = 2\cos^2(x/2) \)
\( 1 - \cos x = 2\sin^2(x/2) \)
Thus:
\( \sqrt{\frac{1 + \cos x}{1 - \cos x}} = \sqrt{\frac{2\cos^2(x/2)}{2\sin^2(x/2)}} = \sqrt{\cot^2(x/2)} = \cot(x/2) \) (as \( x \in (0, \pi/2) \), \( \cot(x/2) > 0 \)).
So, \( T_2 = \cot^{-1}\left(\cot(x/2)\right) = \frac{x}{2} \).
Now, substitute these back into the original expression:
\( \text{LHS} = T_1 - T_2 = \left( \frac{\pi}{4} + \frac{x}{2} \right) - \frac{x}{2} = \frac{\pi}{4} = \text{RHS} \).
Hence proved.
In simple words: Convert the first term into a tan identity and the second term into a cot identity. When you subtract them, the variable terms cancel out, leaving exactly the constant.

Exam Tip: Be careful with signs when substituting \( 1 - \sin x \). Always write it as \( (\cos(x/2) - \sin(x/2))^2 \) rather than the other way around to ensure a positive base in the first quadrant.

Question 5. Prove \( \tan^{-1}\left( \frac{x}{\sqrt{a^2 - x^2}} \right) = \sin^{-1}\left(\frac{x}{a}\right) = \cos^{-1}\left( \frac{\sqrt{a^2 - x^2}}{a} \right) \).
Answer:
Let \( \theta = \tan^{-1}\left( \frac{x}{\sqrt{a^2 - x^2}} \right) \).
Substitute \( x = a\sin\phi \), which implies \( \sin\phi = \frac{x}{a} \) or \( \phi = \sin^{-1}\left(\frac{x}{a}\right) \).
Then, the denominator becomes:
\( \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\phi} = \sqrt{a^2(1 - \sin^2\phi)} = a\cos\phi \).
Substituting these values into the expression:
\( \theta = \tan^{-1}\left( \frac{a\sin\phi}{a\cos\phi} \right) = \tan^{-1}(\tan\phi) = \phi \).
Thus, we have:
\( \theta = \sin^{-1}\left(\frac{x}{a}\right) \).
Now let us express \( \phi \) in terms of \( \cos^{-1} \).
Since \( \cos\phi = \frac{\sqrt{a^2 - x^2}}{a} \), we have:
\( \phi = \cos^{-1}\left( \frac{\sqrt{a^2 - x^2}}{a} \right) \).
Therefore:
\( \tan^{-1}\left( \frac{x}{\sqrt{a^2 - x^2}} \right) = \sin^{-1}\left(\frac{x}{a}\right) = \cos^{-1}\left( \frac{\sqrt{a^2 - x^2}}{a} \right) \).
Hence proved.
In simple words: Substitute \( x = a \sin\phi \) to simplify the fraction inside the inverse tangent, showing that all three inverse functions describe the same angle.

Exam Tip: Substituting trigonometric variables (like \( x = a\sin\phi \)) is a standard technique for simplifying algebraic terms of the form \( \sqrt{a^2-x^2} \).

Question 6. Prove \( \cot^{-1}\left[ 2 \tan\left( \cos^{-1} \frac{8}{17} \right) \right] + \tan^{-1}\left[ 2 \tan\left( \sin^{-1} \frac{8}{17} \right) \right] = \tan^{-1}\left( \frac{300}{161} \right) \).
Answer:
Let us simplify the terms inside the square brackets.
For the first term:
Let \( \alpha = \cos^{-1}\left(\frac{8}{17}\right) \), which means \( \cos\alpha = \frac{8}{17} \).
Using a right-angled triangle, the perpendicular is \( \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \).
Thus, \( \tan\alpha = \frac{15}{8} \).
The first term becomes:
\( \cot^{-1}\left[ 2 \tan\alpha \right] = \cot^{-1}\left[ 2 \cdot \frac{15}{8} \right] = \cot^{-1}\left( \frac{15}{4} \right) \).
Since \( \cot^{-1} u = \tan^{-1}(1/u) \) for positive \( u \), we can write this as:
\( \tan^{-1}\left( \frac{4}{15} \right) \).
For the second term:
Let \( \beta = \sin^{-1}\left(\frac{8}{17}\right) \), which means \( \sin\beta = \frac{8}{17} \).
Using a right-angled triangle, the base is \( \sqrt{17^2 - 8^2} = 15 \).
Thus, \( \tan\beta = \frac{8}{15} \).
The second term becomes:
\( \tan^{-1}\left[ 2 \tan\beta \right] = \tan^{-1}\left[ 2 \cdot \frac{8}{15} \right] = \tan^{-1}\left( \frac{16}{15} \right) \).
Now, add both simplified terms:
\( \text{LHS} = \tan^{-1}\left(\frac{4}{15}\right) + \tan^{-1}\left(\frac{16}{15}\right) \).
Using the identity \( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) (since \( xy = \frac{4}{15} \cdot \frac{16}{15} = \frac{64}{225} < 1 \)):
\( \text{LHS} = \tan^{-1}\left( \frac{\frac{4}{15} + \frac{16}{15}}{1 - \frac{64}{225}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{20}{15}}{\frac{225 - 64}{225}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{4}{3}}{\frac{161}{225}} \right) \)
\( = \tan^{-1}\left( \frac{4}{3} \times \frac{225}{161} \right) \)
\( = \tan^{-1}\left( \frac{4 \times 75}{161} \right) \)
\( = \tan^{-1}\left( \frac{300}{161} \right) = \text{RHS} \).
Hence proved.
In simple words: Convert the inner inverse cosine and sine functions to standard tangent fractions. Multiply them by 2, apply reciprocal rules to make both terms inverse tangents, and then apply the addition formula to combine them.

Exam Tip: Be sure to verify that the product \( xy < 1 \) before applying the standard \( \tan^{-1} x + \tan^{-1} y \) identity, to confirm that you do not need the \( \pi + \tan^{-1} \) form.

Question 7. Prove \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x^2) \).
Answer:
Let \( x^2 = \cos 2\theta \).

\( \implies 2\theta = \cos^{-1}(x^2) \)

\( \implies \theta = \frac{1}{2}\cos^{-1}(x^2) \).
Using half-angle trigonometric formulas, we have:
\( 1 + x^2 = 1 + \cos 2\theta = 2\cos^2\theta \)
\( 1 - x^2 = 1 - \cos 2\theta = 2\sin^2\theta \)
Taking square roots, we get:
\( \sqrt{1 + x^2} = \sqrt{2}\cos\theta \)
\( \sqrt{1 - x^2} = \sqrt{2}\sin\theta \)
Substitute these expressions into the left-hand side (LHS):
\( \text{LHS} = \tan^{-1}\left( \frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta} \right) \)
\( = \tan^{-1}\left( \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \right) \)
Divide the numerator and denominator by \( \cos\theta \):
\( = \tan^{-1}\left( \frac{1 + \tan\theta}{1 - \tan\theta} \right) \)
\( = \tan^{-1}\left( \tan\left( \frac{\pi}{4} + \theta \right) \right) \)
\( = \frac{\pi}{4} + \theta \).
Substitute the value of \( \theta \) back:
\( = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x^2) = \text{RHS} \).
Hence proved.
In simple words: Substituting \( x^2 = \cos 2\theta \) simplifies the square root terms via cosine half-angle formulas, reducing the argument of inverse tangent to a basic tangent expression.

Exam Tip: Remember to always precede the \( \implies \) sign with a line break as done in step 1 - this makes equations very clear and helps score full marks on step presentation.

Question 8. Solve \( \cot^{-1} 2x + \cot^{-1} 3x = \frac{\pi}{4} \).
Answer:
Assuming \( x > 0 \), we can write the complementary relations:
\( \cot^{-1} 2x = \tan^{-1}\left(\frac{1}{2x}\right) \)
\( \cot^{-1} 3x = \tan^{-1}\left(\frac{1}{3x}\right) \)
Thus, the equation becomes:
\( \tan^{-1}\left(\frac{1}{2x}\right) + \tan^{-1}\left(\frac{1}{3x}\right) = \frac{\pi}{4} \).
Applying the sum formula \( \tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \):
\( \tan^{-1}\left( \frac{\frac{1}{2x} + \frac{1}{3x}}{1 - \frac{1}{2x} \cdot \frac{1}{3x}} \right) = \frac{\pi}{4} \)
Taking tangent on both sides:
\( \frac{\frac{3x + 2x}{6x^2}}{\frac{6x^2 - 1}{6x^2}} = \tan\left(\frac{\pi}{4}\right) \)

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

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

\( \implies x = 1 \) or \( x = -\frac{1}{6} \).
Check the validity of solutions:
If \( x = -\frac{1}{6} \), then both \( 2x \) and \( 3x \) are negative, which makes both cotangent terms fall in the interval \( \left(\frac{\pi}{2}, \pi\right) \). Their sum would be greater than \( \pi \), which cannot equal \( \frac{\pi}{4} \).
Thus, \( x = -\frac{1}{6} \) is rejected.
If \( x = 1 \), LHS is \( \cot^{-1}(2) + \cot^{-1}(3) = \tan^{-1}(1/2) + \tan^{-1}(1/3) = \tan^{-1}(1) = \frac{\pi}{4} \). This is valid.
Therefore, the only solution is \( x = 1 \).
In simple words: Convert the cotangent terms into tangent fractions, merge them with the addition rule, solve the resulting quadratic equation, and discard the negative root since it doesn't satisfy the original range.

Exam Tip: Never forget to verify your final roots with the original equation. Discarding extraneous roots is necessary for scoring full marks.

Question 9. Prove that \( \tan^{-1}\left( \frac{m}{n} \right) - \tan^{-1}\left( \frac{m - n}{m + n} \right) = \frac{\pi}{4} \), \( m, n > 0 \).
Answer:
Let us simplify the second term of the left-hand side (LHS):
\( T_2 = \tan^{-1}\left( \frac{m - n}{m + n} \right) \).
Divide both the numerator and the denominator inside the argument by \( n \):
\( T_2 = \tan^{-1}\left( \frac{\frac{m}{n} - 1}{\frac{m}{n} + 1} \right) = \tan^{-1}\left( \frac{\frac{m}{n} - 1}{1 + \left(\frac{m}{n}\right)(1)} \right) \).
Using the identity \( \tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1} x - \tan^{-1} y \), we can rewrite this as:
\( T_2 = \tan^{-1}\left(\frac{m}{n}\right) - \tan^{-1}(1) \).
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\( T_2 = \tan^{-1}\left(\frac{m}{n}\right) - \frac{\pi}{4} \).
Substitute this back into the LHS:
\( \text{LHS} = \tan^{-1}\left(\frac{m}{n}\right) - T_2 \)
\( = \tan^{-1}\left(\frac{m}{n}\right) - \left[ \tan^{-1}\left(\frac{m}{n}\right) - \frac{\pi}{4} \right] \)
\( = \frac{\pi}{4} = \text{RHS} \).
Hence proved.
In simple words: Divide the numerator and denominator of the second term's argument by \( n \), split it using the difference formula for inverse tangent, and subtract to find that the terms cancel out.

Exam Tip: Restructuring the expression inside the inverse tangent by dividing by a variable is a highly elegant approach that saves time and avoids heavy algebraic multiplication.

Question 10. Prove that \( \tan\left[ \frac{1}{2}\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \frac{1}{2}\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) \right] = \frac{x+y}{1-xy} \).
Answer:
Using standard inverse trigonometric identities:
\( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1} x \) for \( |x| \le 1 \)
\( \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) = 2\tan^{-1} y \) for \( y \ge 0 \)
Substitute these two identities into the given expression:
\( \text{LHS} = \tan\left[ \frac{1}{2}\left(2\tan^{-1} x\right) + \frac{1}{2}\left(2\tan^{-1} y\right) \right] \)
\( = \tan\left[ \tan^{-1} x + \tan^{-1} y \right] \)
Using the sum formula \( \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \):
\( = \tan\left[ \tan^{-1}\left( \frac{x+y}{1-xy} \right) \right] \)
\( = \frac{x+y}{1-xy} = \text{RHS} \).
Hence proved.
In simple words: Replace the inverse sine and inverse cosine terms with their double-angle tangent equivalents, divide by 2, and use the tangent sum identity to obtain the desired result.

Exam Tip: Always state the domains under which the identities \( \sin^{-1}(\frac{2x}{1+x^2}) = 2\tan^{-1}x \) and \( \cos^{-1}(\frac{1-y^2}{1+y^2}) = 2\tan^{-1}y \) are valid, showing complete conceptual understanding.

Question 11. Solve for x, \( \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) + \frac{1}{2}\tan^{-1}\left(\frac{-2x}{1-x^2}\right) = \frac{2\pi}{3} \).
Answer:
Let \( x = \tan\theta \) where \( \theta \in \left(0, \frac{\pi}{2}\right) \).
First term:
\( \frac{x^2-1}{x^2+1} = -\left(\frac{1-x^2}{1+x^2}\right) = -\cos 2\theta \).
Thus, we have:
\( \cos^{-1}\left( -\cos 2\theta \right) = \pi - \cos^{-1}(\cos 2\theta) = \pi - 2\theta \) (assuming \( 2\theta \in (0, \pi) \)).
Second term:
\( \frac{-2x}{1-x^2} = \frac{2x}{x^2-1} = -\tan 2\theta \).
Thus, assuming \( 2\theta \in \left(0, \frac{\pi}{2}\right) \), we get:
\( \frac{1}{2}\tan^{-1}\left(-\tan 2\theta\right) = \frac{1}{2}(-2\theta) = -\theta \).
Substituting these back into the equation:
\( (\pi - 2\theta) - \theta = \frac{2\pi}{3} \)

\( \implies \pi - 3\theta = \frac{2\pi}{3} \)

\( \implies 3\theta = \pi - \frac{2\pi}{3} = \frac{\pi}{3} \)

\( \implies \theta = \frac{\pi}{9} \).
Thus, \( x = \tan\left(\frac{\pi}{9}\right) \).

Alternatively, if the intended equation from standard textbook patterns (corrected for typos in the term coefficient) yields: \( \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) + \tan^{-1}\left(\frac{-2x}{1-x^2}\right) = \frac{2\pi}{3} \),
we find:
\( (\pi - 2\theta) - 2\theta = \frac{2\pi}{3} \)

\( \implies \pi - 4\theta = \frac{2\pi}{3} \)

\( \implies 4\theta = \frac{\pi}{3} \)

\( \implies \theta = \frac{\pi}{12} \).
In this case, the solution is:
\( x = \tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3} \).
In simple words: Use substitution \( x = \tan\theta \), simplify using double-angle formulas for cosine and tangent, solve for \( \theta \), and convert back to find \( x \).

Exam Tip: If you face an equation with potential textbook typos, write out the systematic steps of substitution clearly; this ensures you receive maximum step marks regardless of the final numerical value.

Question 12. 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:
Group the four terms into two pairs:
\( \text{LHS} = \left[ \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) \right] + \left[ \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) \right] \).
Apply the sum formula to each group:
For the first group:
\( \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3}\cdot\frac{1}{5}} \right) = \tan^{-1}\left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan^{-1}\left(\frac{8}{14}\right) = \tan^{-1}\left(\frac{4}{7}\right) \).
For the second group:
\( \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7}\cdot\frac{1}{8}} \right) = \tan^{-1}\left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan^{-1}\left(\frac{15}{55}\right) = \tan^{-1}\left(\frac{3}{11}\right) \).
Now add the two resulting terms:
\( \text{LHS} = \tan^{-1}\left(\frac{4}{7}\right) + \tan^{-1}\left(\frac{3}{11}\right) \)
\( = \tan^{-1}\left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7}\cdot\frac{3}{11}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{44 + 21}{77}}{\frac{77 - 12}{77}} \right) \)
\( = \tan^{-1}\left( \frac{65/77}{65/77} \right) \)
\( = \tan^{-1}(1) = \frac{\pi}{4} = \text{RHS} \).
Hence proved.
In simple words: Combine the four inverse tangents in pairs, simplify the fractions, and then combine the two resulting terms to get inverse tangent of 1, which equals \( \pi/4 \).

Exam Tip: Grouping terms strategically (e.g. pairing 1/3 with 1/5 and 1/7 with 1/8) keeps the common denominators relatively small and arithmetic easy to handle.

Question 13. Solve for x, \( \tan(\cos^{-1} x) = \sin(\tan^{-1} 2) \); \( x > 0 \).
Answer:
Let us evaluate the right-hand side (RHS) first.
Let \( \theta = \tan^{-1} 2 \), which means \( \tan\theta = 2 \).
In a right-angled triangle, the perpendicular is 2 and the base is 1, so the hypotenuse is \( \sqrt{1^2 + 2^2} = \sqrt{5} \).
Thus, \( \sin\theta = \frac{2}{\sqrt{5}} \).
Therefore, \( \text{RHS} = \frac{2}{\sqrt{5}} \).
Now, let us simplify the left-hand side (LHS).
Let \( \phi = \cos^{-1} x \), which means \( \cos\phi = x \).
In a right-angled triangle, the base is \( x \) and the hypotenuse is 1, so the perpendicular is \( \sqrt{1 - x^2} \).
Thus, \( \tan\phi = \frac{\sqrt{1 - x^2}}{x} \) (since \( x > 0 \)).
Therefore, \( \text{LHS} = \frac{\sqrt{1 - x^2}}{x} \).
Now equate LHS and RHS:
\( \frac{\sqrt{1 - x^2}}{x} = \frac{2}{\sqrt{5}} \).
Squaring both sides:
\( \frac{1 - x^2}{x^2} = \frac{4}{5} \)

\( \implies 5(1 - x^2) = 4x^2 \)

\( \implies 5 - 5x^2 = 4x^2 \)

\( \implies 9x^2 = 5 \)

\( \implies x^2 = \frac{5}{9} \).
Since \( x > 0 \), we take the positive square root:
\( x = \frac{\sqrt{5}}{3} \).
In simple words: Represent both sides using triangle geometry to convert the nested functions into simple algebraic forms, then square and solve for \( x \).

Exam Tip: When using the right-angled triangle method, draw a quick small triangle with labeled sides on your answer sheet - this makes your steps visually clear to the examiner.

Question 14. Prove that \( 2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43} \).
Answer:
First, convert the first term using the double-angle identity \( 2\tan^{-1} u = \tan^{-1}\left(\frac{2u}{1-u^2}\right) \):
\( 2\tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left( \frac{2(1/5)}{1 - 1/25} \right) = \tan^{-1}\left( \frac{2/5}{24/25} \right) = \tan^{-1}\left( \frac{2}{5} \times \frac{25}{24} \right) = \tan^{-1}\left(\frac{5}{12}\right) \).
Substitute this back into the LHS:
\( \text{LHS} = \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{1}{4}\right) \).
Applying the sum formula:
\( \text{LHS} = \tan^{-1}\left( \frac{\frac{5}{12} + \frac{1}{4}}{1 - \frac{5}{12}\cdot\frac{1}{4}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{5 + 3}{12}}{1 - \frac{5}{48}} \right) \)
\( = \tan^{-1}\left( \frac{8/12}{43/48} \right) \)
\( = \tan^{-1}\left( \frac{2}{3} \times \frac{48}{43} \right) \)
\( = \tan^{-1}\left(\frac{32}{43}\right) = \text{RHS} \).
Hence proved.
In simple words: Use the double angle formula to simplify the first term, then merge the two terms together using the standard inverse tangent addition formula.

Exam Tip: Make sure you simplify fractions step-by-step rather than trying to calculate the final values in a single line - this prevents basic arithmetic slips.

Question 15. Evaluate \( \tan\left[ \frac{1}{2}\cos^{-1}\left( \frac{3}{\sqrt{11}} \right) \right] \).
Answer:
Let \( \theta = \cos^{-1}\left(\frac{3}{\sqrt{11}}\right) \), which means \( \cos\theta = \frac{3}{\sqrt{11}} \).
We need to evaluate \( \tan\left(\frac{\theta}{2}\right) \).
Using the half-angle formula for tangent:
\( \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{1 + \cos\theta} \)
Substitute the value of \( \cos\theta \):
\( \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \frac{3}{\sqrt{11}}}{1 + \frac{3}{\sqrt{11}}} = \frac{\sqrt{11} - 3}{\sqrt{11} + 3} \).
To simplify, we can rationalize the denominator:
\( \frac{\sqrt{11} - 3}{\sqrt{11} + 3} \times \frac{\sqrt{11} - 3}{\sqrt{11} - 3} = \frac{(\sqrt{11} - 3)^2}{11 - 9} = \frac{(\sqrt{11} - 3)^2}{2} \).
Now, take the positive square root (since \( \theta/2 \) is in the first quadrant):
\( \tan\left(\frac{\theta}{2}\right) = \frac{\sqrt{11} - 3}{\sqrt{2}} \).
Alternatively, expressing the final fraction without rationalizing, as given in the textbook keys:
\( \tan\left(\frac{\theta}{2}\right) = \frac{\sqrt{\sqrt{11} - 3}}{\sqrt{3 + \sqrt{11}}} \).
Both versions are mathematically equivalent.
In simple words: Substitute \( \cos\theta = 3/\sqrt{11} \) into the half-angle formula for tangent, and simplify the square root of the resulting fraction.

Exam Tip: Rationalizing your final expression is a great habit; even if the answer key writes it in a non-rationalized form, standard markings award full marks for either correct form.

Question 16. Prove that \( \tan^{-1}\left( \frac{a \cos x - b \sin x}{b \cos x + a \sin x} \right) = \tan^{-1}\left(\frac{a}{b}\right) - x \).
Answer:
Let us divide both the numerator and the denominator inside the argument by \( b \cos x \):
\( \text{LHS} = \tan^{-1}\left( \frac{\frac{a \cos x}{b \cos x} - \frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x} + \frac{a \sin x}{b \cos x}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{a}{b} - \tan x}{1 + \left(\frac{a}{b}\right)\tan x} \right) \).
Applying the difference identity \( \tan^{-1}\left(\frac{u - v}{1 + uv}\right) = \tan^{-1} u - \tan^{-1} v \) with \( u = \frac{a}{b} \) and \( v = \tan x \):
\( = \tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}(\tan x) \)
\( = \tan^{-1}\left(\frac{a}{b}\right) - x = \text{RHS} \).
Hence proved.
In simple words: Divide the numerator and denominator by \( b \cos x \), rearrange the terms to match the tangent subtraction formula, and directly split them to get the final result.

Exam Tip: Look out for terms in the denominator of the form \( 1 + \ldots \) or a constant that can be scaled to 1. This is a standard signal to divide by that term to make it match the \( \tan^{-1} \) identity format.

Question 17. Prove that \( \cot\left\{ \tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) \right\} + \cos^{-1}\left(1 - 2x^2\right) + \cos^{-1}\left(2x^2 - 1\right) = \pi \), \( x > 0 \).
Answer:
Let us evaluate the terms step-by-step.
First term: \( T_1 = \cot\left\{ \tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) \right\} \)
Since \( x > 0 \), we know \( \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x \).
Using the identity \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \), we have:
\( T_1 = \cot\left( \frac{\pi}{2} \right) = 0 \).
Second and third terms: \( T_2 = \cos^{-1}\left(1 - 2x^2\right) + \cos^{-1}\left(2x^2 - 1\right) \)
We can write \( 2x^2 - 1 = -\left(1 - 2x^2\right) \).
Let \( u = 1 - 2x^2 \). Then:
\( T_2 = \cos^{-1}(u) + \cos^{-1}(-u) \).
Using the negative-argument identity for cosine, \( \cos^{-1}(-u) = \pi - \cos^{-1} u \):
\( T_2 = \cos^{-1} u + \left( \pi - \cos^{-1} u \right) = \pi \).
Adding both results:
\( \text{LHS} = T_1 + T_2 = 0 + \pi = \pi = \text{RHS} \).
Hence proved.
In simple words: Simplify the cotangent term using complementary angles to get 0, and simplify the two cosine terms by rewriting one argument as the negative of the other, which sums to \( \pi \).

Exam Tip: Do not get distracted trying to expand \( \cos^{-1}(1-2x^2) \) as a double-angle formula - recognizing that \( 2x^2-1 \) is simply the negative of \( 1-2x^2 \) makes this a very simple 2-step question.

Question 18. Prove that \( \tan^{-1}\left( \frac{a - b}{1 + ab} \right) + \tan^{-1}\left( \frac{b - c}{1 + bc} \right) + \tan^{-1}\left( \frac{c - a}{1 + ca} \right) = 0 \), where \( a, b, c > 0 \).
Answer:
Applying the difference formula \( \tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1} x - \tan^{-1} y \) to each of the three terms in the LHS:
\( \tan^{-1}\left( \frac{a - b}{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b \)
\( \tan^{-1}\left( \frac{b - c}{1 + bc} \right) = \tan^{-1} b - \tan^{-1} c \)
\( \tan^{-1}\left( \frac{c - a}{1 + ca} \right) = \tan^{-1} c - \tan^{-1} a \)
Add these three equations together:
\( \text{LHS} = \left(\tan^{-1} a - \tan^{-1} b\right) + \left(\tan^{-1} b - \tan^{-1} c\right) + \left(\tan^{-1} c - \tan^{-1} a\right) \)
\( = \tan^{-1} a - \tan^{-1} a - \tan^{-1} b + \tan^{-1} b - \tan^{-1} c + \tan^{-1} c = 0 = \text{RHS} \).
Hence proved.
In simple words: Split each of the three terms into simple subtraction expressions. When you write them all out, every single term cancels out perfectly, leaving zero.

Exam Tip: This is a very common cyclic identity question. Writing out the expansion steps clearly in columns can make the cancellations easy to follow.

Question 19. Solve for x, \( 2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x) \).
Answer:
Apply the double-angle identity \( 2\tan^{-1} u = \tan^{-1}\left(\frac{2u}{1-u^2}\right) \) to the LHS:
\( 2\tan^{-1}(\cos x) = \tan^{-1}\left( \frac{2\cos x}{1 - \cos^2 x} \right) = \tan^{-1}\left( \frac{2\cos x}{\sin^2 x} \right) \).
Substitute this back into the equation:
\( \tan^{-1}\left( \frac{2\cos x}{\sin^2 x} \right) = \tan^{-1}(2\csc x) \).
Equating the arguments:
\( \frac{2\cos x}{\sin^2 x} = 2\csc x \)

\( \implies \frac{\cos x}{\sin^2 x} = \frac{1}{\sin x} \).
Since \( \csc x \) must be defined, \( \sin x \ne 0 \). Thus, we can multiply both sides by \( \sin x \):
\( \frac{\cos x}{\sin x} = 1 \)

\( \implies \cot x = 1 \)

\( \implies x = \frac{\pi}{4} \).
In simple words: Use the double angle formula to turn the left side into a single tangent inverse, equate the inner parts of both sides, simplify the trigonometry, and solve for \( x \).

Exam Tip: Always state that \( \sin x \neq 0 \) explicitly when multiplying both sides by \( \sin x \) to show that no roots are lost or invalid divisions made.

Question 20. Express \( \sin^{-1}\left( x\sqrt{1 - x} - \sqrt{x}\sqrt{1 - x^2} \right) \) in simplest form.
Answer:
Let \( x = \sin A \) and \( \sqrt{x} = \sin B \).
This implies:
\( A = \sin^{-1} x \)
\( B = \sin^{-1}\sqrt{x} \).
Now express the square roots in terms of cosine:
\( \sqrt{1 - x^2} = \cos A \)
\( \sqrt{1 - x} = \sqrt{1 - \left(\sqrt{x}\right)^2} = \cos B \).
Substitute these expressions into the given argument:
\( \sin^{-1}\left( \sin A \cos B - \cos A \sin B \right) \).
Using the subtraction identity \( \sin(A - B) = \sin A \cos B - \cos A \sin B \):
\( = \sin^{-1}\left( \sin(A - B) \right) \)
\( = A - B \).
Substitute back the values of \( A \) and \( B \):
\( = \sin^{-1} x - \sin^{-1}\sqrt{x} \).
In simple words: Substitute variables for both \( x \) and \( \sqrt{x} \) to make the inner term look like a sine difference identity, then cancel the outer inverse sine.

Exam Tip: Recognizing the structure \( x\sqrt{1-y^2} - y\sqrt{1-x^2} \) inside an inverse sine is a classic clue to use the sine subtraction formula.

Question 21. If \( \tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi \), then prove that \( a + b + c = abc \).
Answer:
Let:
\( \tan^{-1} a = \alpha \implies a = \tan\alpha \)
\( \tan^{-1} b = \beta \implies b = \tan\beta \)
\( \tan^{-1} c = \gamma \implies c = \tan\gamma \).
We are given:
\( \alpha + \beta + \gamma = \pi \)

\( \implies \alpha + \beta = \pi - \gamma \).
Taking tangent on both sides:
\( \tan(\alpha + \beta) = \tan(\pi - \gamma) \)
Using the sum formula for tangent and knowing that \( \tan(\pi - \gamma) = -\tan\gamma \):
\( \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = -\tan\gamma \).
Substitute back the algebraic values \( a, b, c \):
\( \frac{a + b}{1 - ab} = -c \)

\( \implies a + b = -c(1 - ab) \)

\( \implies a + b = -c + abc \)

\( \implies a + b + c = abc \).
Hence proved.
In simple words: Represent the inverse tangents as angles summing to \( \pi \), write the tangent of their sum, substitute the original values back, and rearrange the equation.

Exam Tip: When taking tangent of both sides, don't forget the quadrant change: \( \tan(\pi - \gamma) = -\tan\gamma \). Dropping this minus sign is the most common error.

Question 22. If \( \sin^{-1} x > \cos^{-1} x \), then \( x \) belongs to which interval?
Answer:
Using the complementary angle identity, we have:
\( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \)

\( \implies \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \).
Substitute this into the given inequality:
\( \sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x \)

\( \implies 2\sin^{-1} x > \frac{\pi}{2} \)

\( \implies \sin^{-1} x > \frac{\pi}{4} \).
Since the inverse sine function is strictly increasing on its domain \( [-1, 1] \), we can apply sine to both sides without changing the inequality direction:
\( x > \sin\left(\frac{\pi}{4}\right) \)

\( \implies x > \frac{1}{\sqrt{2}} \).
Combining this with the absolute upper boundary of the domain of inverse sine, which is 1, we get:
\( x \in \left( \frac{1}{\sqrt{2}}, 1 \right] \).
In simple words: Replace inverse cosine with \( \pi/2 \) minus inverse sine, rearrange to find that inverse sine must be greater than \( \pi/4 \), and state the values up to the domain limit of 1.

Exam Tip: Always specify the upper limit of the domain (1) in your final interval; leaving it open-ended as \( x > \frac{1}{\sqrt{2}} \) will lose marks for precision.

Please click the below link to access CBSE Class 12 Mathematics Inverse Trigonometric Functions Assignment Set A