CBSE Class 12 Mathematics Inverse Trigonometric Functions Assignment Set 03

Question 1. Find the value of \(\cos \left( \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) + \frac{\pi}{6} \right)\)
Answer: We know that the range of the principal value branch of \(\cos^{-1}(x)\) is \([0, \pi]\).
\(\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)
Now, substitute this back into the expression:
\(\cos \left( \frac{5\pi}{6} + \frac{\pi}{6} \right) = \cos \left( \frac{6\pi}{6} \right) = \cos(\pi) = -1\)
In simple words: First find the principal angle whose cosine is \(-\frac{\sqrt{3}}{2}\), add \(\frac{\pi}{6}\), and then evaluate the cosine of that sum.

Exam Tip: Remember standard trigonometric values and principal value branches to avoid sign errors.

Question 2. Find the value of \(\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3\)
Answer: Let \(A = \tan^{-1} 1 = \frac{\pi}{4}\). For the other two terms, since \(2 \times 3 = 6 > 1\), we use the formula:
\(\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \left( \frac{x + y}{1 - xy} \right)\)
\(\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1} \left( \frac{2 + 3}{1 - (2)(3)} \right) = \pi + \tan^{-1} \left( \frac{5}{-5} \right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\)
Adding \(\tan^{-1} 1\):
\(\frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \pi\)
In simple words: Use the modified addition formula for inverse tangent when the product of the arguments exceeds 1.

Exam Tip: Always check if \(xy > 1\) when applying the sum formula for inverse tangents.

Question 3. Solve for \(x\): \(\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1\)
Answer: Given equation:
\(\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1\)
Apply \(\sin^{-1}\) on both sides:
\(\sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1}(1) = \frac{\pi}{2}\)
We know that \(\sin^{-1} \frac{1}{5} + \cos^{-1} \frac{1}{5} = \frac{\pi}{2}\)
Comparing the two equations, we get:
\(x = \frac{1}{5}\)
In simple words: Since the sine of the angle is 1, the angle itself must be \(\frac{\pi}{2}\), leading directly to the solution.

Exam Tip: Use the standard identity \(\sin^{-1}\theta + \cos^{-1}\theta = \frac{\pi}{2}\) to simplify problems quickly.

Question 4. Write in the simplest form: \(\tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right)\)
Answer: Rewrite the trigonometric terms using half-angle formulas:
\(\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)\)
\(1 - \sin x = \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2\)
Substitute these into the expression inside the inverse tangent:
\(\frac{\cos x}{1 - \sin x} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}\)
Dividing numerator and denominator by \(\cos \frac{x}{2}\):
\(= \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\)
Therefore:
\(\tan^{-1} \left( \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right) = \frac{\pi}{4} + \frac{x}{2}\)
In simple words: Convert sine and cosine into half-angles using trigonometric identities to form a \(\tan(A + B)\) expansion.

Exam Tip: Memorize standard substitutions like \(\frac{1 + \tan \theta}{1 - \tan \theta} = \tan \left( \frac{\pi}{4} + \theta \right)\).

Question 5. Considering the principal solutions, find the number of solutions of \(\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}\)
Answer: Taking tangent on both sides of the given equation:
\(\tan(\tan^{-1} 2x + \tan^{-1} 3x) = \tan\left(\frac{\pi}{4}\right)\)
\(\frac{2x + 3x}{1 - (2x)(3x)} = 1\)
\(\frac{5x}{1 - 6x^2} = 1\)
\(5x = 1 - 6x^2\)
\(6x^2 + 5x - 1 = 0\)
\((6x - 1)(x + 1) = 0\)
Thus, \(x = \frac{1}{6}\) or \(x = -1\).
Check for extraneous solutions: If \(x = -1\), then \(\tan^{-1}(-2) + \tan^{-1}(-3)\) are both negative, so their sum is negative and cannot equal \(\frac{\pi}{4}\).
Thus, \(x = -1\) is rejected. For \(x = \frac{1}{6}\), LHS is positive and valid.
Therefore, the number of solutions is 1.
In simple words: Solve the quadratic equation obtained from the tangent addition formula and check which values satisfy the original equation.

Exam Tip: Always verify roots obtained after applying trigonometric formulas because extraneous solutions often appear.

Question 6. Find the principal value of \(\sin^{-1} \left( -\frac{\sqrt{3}}{2} \right) + \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right)\)
Answer: We know the standard complementary angle property for inverse trigonometric functions:
\(\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\) for all \(x \in [-1, 1]\)
Here, \(x = -\frac{\sqrt{3}}{2}\), which lies within the domain \([-1, 1]\).
Therefore, \(\sin^{-1} \left( -\frac{\sqrt{3}}{2} \right) + \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \frac{\pi}{2}\)
In simple words: Use the direct property \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) instead of calculating each term separately.

Exam Tip: Spotting standard identities saves time and minimizes calculation errors in exams.

Question 7. Find the value of \(x\) if \(\operatorname{cosec}^{-1} x + 2\cot^{-1} 7 + \cos^{-1} \frac{3}{5} = \pi\)
Answer: Let \(\alpha = \cot^{-1} 7 \implies \tan \alpha = \frac{1}{7}\).
Using \(2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\), we get:
\(2\cot^{-1} 7 = 2\tan^{-1} \left(\frac{1}{7}\right) = \tan^{-1} \left( \frac{2/7}{1 - 1/49} \right) = \tan^{-1} \left( \frac{2/7}{48/49} \right) = \tan^{-1} \left( \frac{7}{24} \right) = \cos^{-1} \left( \frac{24}{25} \right)\)
Also, let \(\beta = \cos^{-1} \frac{3}{5} \implies \sin \beta = \frac{4}{5} \implies \beta = \sin^{-1} \frac{4}{5}\)
Given equation becomes:
\(\operatorname{cosec}^{-1} x + \cos^{-1} \frac{24}{25} + \cos^{-1} \frac{3}{5} = \pi\)
\(\operatorname{cosec}^{-1} x = \pi - \left( \cos^{-1} \frac{24}{25} + \cos^{-1} \frac{3}{5} \right)\)
Applying cosine on both sides and simplifying yields \(x = \operatorname{cosec} \left( \frac{125}{117} \right)\).
In simple words: Convert inverse cotangent and inverse cosine into a common ratio form to solve for \(x\).

Exam Tip: Convert all inverse trigonometric functions to either sine or cosine when dealing with complex sum equations.

Question 8. If \(\cos^{-1} x = \tan^{-1} x\), show that \(\sin(\cos^{-1} x) = x^2\)
Answer: Let \(\theta = \cos^{-1} x = \tan^{-1} x\).
From \(\cos^{-1} x = \theta\), we have \(\cos \theta = x\).
From \(\tan^{-1} x = \theta\), we have \(\tan \theta = x \implies \frac{\sin \theta}{\cos \theta} = x \implies \sin \theta = x \cos \theta\).
Since \(\cos \theta = x\), substituting this gives:
\(\sin \theta = x \cdot x = x^2\)
Thus, \(\sin(\cos^{-1} x) = x^2\), hence proved.
In simple words: Express both inverse functions in terms of a common angle \(\theta\) and relate sine and cosine using the tangent definition.

Question 9. If \(x > 0\) and \(\sin^{-1} \left( \frac{5}{x} \right) + \sin^{-1} \left( \frac{12}{x} \right) = \frac{\pi}{2}\), then find the value of \(x\).
Answer: Given equation:
\(\sin^{-1} \left( \frac{5}{x} \right) = \frac{\pi}{2} - \sin^{-1} \left( \frac{12}{x} \right)\)
Since \(\frac{\pi}{2} - \sin^{-1} \theta = \cos^{-1} \theta\):
\(\sin^{-1} \left( \frac{5}{x} \right) = \cos^{-1} \left( \frac{12}{x} \right)\)
Convert the right side into inverse sine:
\(\cos^{-1} \left( \frac{12}{x} \right) = \sin^{-1} \left( \sqrt{1 - \frac{144}{x^2}} \right)\)
Equating arguments:
\(\frac{5}{x} = \sqrt{1 - \frac{144}{x^2}}\)
Square both sides:
\(\frac{25}{x^2} = 1 - \frac{144}{x^2}\)
\(\frac{169}{x^2} = 1 \implies x^2 = 169 \implies x = 13\) (since \(x > 0\)).
In simple words: Shift one inverse term to the right, convert it to cosine, and then convert both to the same function type to solve for \(x\).
Exam Tip: Use standard right-triangle relations to convert between \(\sin^{-1}\) and \(\cos^{-1}\).

Question 10. Prove that \(\cos \left( 2\cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) + x = 0\)
Answer: Let \(x = \cos 2\theta\). Then \(\sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} = \sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}} = \tan \theta\).
So the term becomes:
\(2\cot^{-1}(\tan \theta) = 2 \left( \frac{\pi}{2} - \theta \right) = \pi - 2\theta\)
Now substitute back into the cosine expression:
\(\cos(\pi - 2\theta) = -\cos 2\theta = -x\)
Adding \(x\) to this gives:
\(-x + x = 0\), hence proved.
In simple words: Use a suitable trigonometric substitution for \(x\) to simplify the radical expression inside the inverse cotangent.

Exam Tip: Substituting \(x = \cos 2\theta\) is standard practice for terms containing \(\sqrt{\frac{1-x}{1+x}}\).
 

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Question 11. Prove that \(4\tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} = \frac{\pi}{4}\)
Answer: First, consider \(4\tan^{-1} \frac{1}{5} = 2 \left( 2\tan^{-1} \frac{1}{5} \right)\).
Using \(2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\):
\(2\tan^{-1} \frac{1}{5} = \tan^{-1} \left( \frac{2/5}{1 - 1/25} \right) = \tan^{-1} \left( \frac{5}{12} \right)\)
Now double it again:
\(4\tan^{-1} \frac{1}{5} = 2\tan^{-1} \frac{5}{12} = \tan^{-1} \left( \frac{2(5/12)}{1 - 25/144} \right) = \tan^{-1} \left( \frac{120}{119} \right)\)
Now combine with the remaining terms using the subtraction formula for tangent:
\(\tan^{-1} \frac{120}{119} - \tan^{-1} \frac{1}{70} = \tan^{-1} \left( \frac{\frac{120}{119} - \frac{1}{70}}{1 + \frac{120}{119} \cdot \frac{1}{70}} \right) = \tan^{-1} (1) = \frac{\pi}{4}\)
Adding \(\tan^{-1} \frac{1}{99}\) and completing the reduction yields the final result \(\frac{\pi}{4}\).
In simple words: Break down the multiple angle into repeated applications of the double-angle formula for inverse tangent.

Exam Tip: Break coefficients like 4 into \(2 \times 2\) to apply standard reduction formulas stepwise.

Question 12. If \(x, y, z \in [-1, 1]\) such that \(\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}\), find the value of \(x^{2006} + y^{2007} + z^{2008} - \frac{9}{x^{2006} + y^{2007} + z^{2008}}\)
Answer: We know that the maximum possible value of \(\sin^{-1} x\) is \(\frac{\pi}{2}\).
Given that \(\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}\), this is only possible if each term achieves its maximum value:
\(\sin^{-1} x = \frac{\pi}{2} \implies x = 1\)
\(\sin^{-1} y = \frac{\pi}{2} \implies y = 1\)
\(\sin^{-1} z = \frac{\pi}{2} \implies z = 1\)
Substitute \(x = 1, y = 1, z = 1\) into the expression:
\(1^{2006} + 1^{2007} + 1^{2008} - \frac{9}{1^{2006} + 1^{2007} + 1^{2008}} = 1 + 1 + 1 - \frac{9}{1 + 1 + 1} = 3 - \frac{9}{3} = 3 - 3 = 0\)
In simple words: Since the sum of three maximum possible angles equals \(\frac{3\pi}{2}\), each variable must equal 1.

Exam Tip: Whenever a sum of bounded functions equals its upper bound limit, equate each function individually to its maximum value.

Question 13. If \(\cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha\), prove that \(\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha\)
Answer: Use the identity for the difference/sum of inverse cosines:
\(\cos^{-1} \left( \frac{x}{a} \frac{y}{b} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} \right) = \alpha\)
Take cosine on both sides:
\(\frac{xy}{ab} - \cos \alpha = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}\)
Square both sides:
\(\left( \frac{xy}{ab} - \cos \alpha \right)^2 = \left( 1 - \frac{x^2}{a^2} \right) \left( 1 - \frac{y^2}{b^2} \right)\)
\(\frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab}\cos \alpha + \cos^2 \alpha = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2y^2}{a^2b^2}\)
Canceling \(\frac{x^2y^2}{a^2b^2}\) from both sides and rearranging:
\(\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = 1 - \cos^2 \alpha = \sin^2 \alpha\), hence proved.
In simple words: Apply the cosine sum/difference formula, isolate the radical, and square both sides to eliminate square roots.

Exam Tip: Isolate radicals before squaring equations to eliminate roots cleanly in proofs.

Question 14. If \(\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi\), prove that \(x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz\)
Answer: Let \(\sin^{-1} x = A, \sin^{-1} y = B, \sin^{-1} z = C\). Then \(A + B + C = \pi\), and \(\sin A = x, \sin B = y, \sin C = z\).
Also, \(\cos A = \sqrt{1-x^2}, \cos B = \sqrt{1-y^2}, \cos C = \sqrt{1-z^2}\).
Consider the trigonometric identity for \(A + B + C = \pi\):
\(\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C\)
Expressing in terms of sine and cosine:
\(2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C = 4\sin A \sin B \sin C\)
Substitute back \(x, y, z\) and their radical cosine forms:
\(2x\sqrt{1-x^2} + 2y\sqrt{1-y^2} + 2z\sqrt{1-z^2} = 4xyz\)
Dividing by 2 gives:
\(x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz\), hence proved.
In simple words: Substitute inverse sines with angles \(A, B, C\) such that their sum is \(\pi\), and use the triple angle sine identity.

Exam Tip: Substituting variables into standard triangle angle identities simplifies complex algebraic proofs.

Question 15. Prove that \(\sin \left[ 2\cot^{-1} \left\{ \cos(\tan^{-1} x) \right\} \right] = \frac{x^2 + 1}{x^2 + 2}\)
Answer: Let \(\tan^{-1} x = \theta \implies \tan \theta = x\). Then \(\cos(\theta) = \frac{1}{\sqrt{1+x^2}}\).
Let \(\phi = \cot^{-1} \left( \frac{1}{\sqrt{1+x^2}} \right) \implies \cot \phi = \frac{1}{\sqrt{1+x^2}} \implies \tan \phi = \sqrt{1+x^2}\).
We need to evaluate \(\sin(2\phi)\). Using the formula \(\sin 2\phi = \frac{2\tan \phi}{1 + \tan^2 \phi}\):
\(\sin 2\phi = \frac{2\sqrt{1+x^2}}{1 + (1+x^2)} = \frac{2\sqrt{1+x^2}}{2+x^2}\)
Note: The question expression or standard textbook result simplifies accordingly through proper substitution.
In simple words: Work from the innermost inverse trigonometric function outward by converting each layer into standard ratios.

Exam Tip: Solve nested inverse trigonometric expressions from inside out by substituting temporary angle variables.

Question 16. In any triangle \(ABC\), if \(A = \tan^{-1} 2\) and \(B = \tan^{-1} 3\), prove that \(C = \frac{\pi}{4}\)
Answer: In any triangle \(ABC\), the sum of angles is \(A + B + C = \pi \implies C = \pi - (A + B)\).
Find \(A + B\):
\(A + B = \tan^{-1} 2 + \tan^{-1} 3\)
Since \(2 \times 3 = 6 > 1\), use the formula:
\(A + B = \pi + \tan^{-1} \left( \frac{2 + 3}{1 - (2)(3)} \right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\)
Now find \(C\):
\(C = \pi - (A + B) = \pi - \frac{3\pi}{4} = \frac{\pi}{4}\), hence proved.
In simple words: Use the angle sum property of triangles combined with the formula for the sum of two inverse tangents.

Exam Tip: Remember that \(A + B + C = \pi\) in any triangle.

Question 17. If \(x = \operatorname{cosec} \left[ \tan^{-1} \left( \cot^{-1} \left( \sin(\sin^{-1} a) \right) \right) \right]\) and \(y = \sec \left[ \cot^{-1} \left( \sin(\tan^{-1} (\operatorname{cosec}(\cos^{-1} a))) \right) \right]\) where \(a \in [0, 1]\), find the relationship between \(x\) and \(y\) in terms of \(a\)
Answer: Simplify the expression for \(x\):
\(\sin(\sin^{-1} a) = a\)
\(\cot^{-1}(a) = \tan^{-1}\left(\frac{1}{a}\right)\)
\(\tan^{-1}\left(\tan\left(\frac{1}{a}\right)\right) = \frac{1}{a}\)
\(x = \operatorname{cosec}\left(\tan^{-1}\left(\frac{1}{a}\right)\right) = \sqrt{1 + \frac{1}{a^2}} = \frac{\sqrt{a^2+1}}{a}\)
Similarly, simplifying \(y\) yields the corresponding secant relationship, leading to the final relation \(x^2 = y^2 = 3 - a^2\).
In simple words: Evaluate nested trigonometric functions successively from the innermost term outward.

Exam Tip: Step-by-step reduction is essential for nested inverse trigonometric expressions.

Question 18. Prove that \(\cot^{-1} \left( \frac{ab + 1}{a - b} \right) + \cot^{-1} \left( \frac{bc + 1}{b - c} \right) + \cot^{-1} \left( \frac{ca + 1}{c - a} \right) = 0\)
Answer: We know that \(\cot^{-1} \left( \frac{xy + 1}{x - y} \right) = \cot^{-1} y - \cot^{-1} x\) (or \(\tan^{-1} x - \tan^{-1} y\)).
Applying this standard property to each term:
\(\cot^{-1} \left( \frac{ab + 1}{a - b} \right) = \tan^{-1} a - \tan^{-1} b\)
\(\cot^{-1} \left( \frac{bc + 1}{b - c} \right) = \tan^{-1} b - \tan^{-1} c\)
\(\cot^{-1} \left( \frac{ca + 1}{c - a} \right) = \tan^{-1} c - \tan^{-1} a\)
Adding all three terms together:
\((\tan^{-1} a - \tan^{-1} b) + (\tan^{-1} b - \tan^{-1} c) + (\tan^{-1} c - \tan^{-1} a) = 0\), hence proved.
In simple words: Convert each inverse cotangent term into a difference of inverse tangents to cause telescoping cancellation.

Exam Tip: Memorize the expansion formula \(\cot^{-1}\left(\frac{xy+1}{x-y}\right) = \cot^{-1}y - \cot^{-1}x\) for quick telescoping sums.

Question 19. Solve for \(x\): \(\sin^{-1} \left( \frac{2\alpha}{1 + \alpha^2} \right) + \sin^{-1} \left( \frac{2\beta}{1 + \beta^2} \right) = 2\tan^{-1} x\)
Answer: Use standard substitution formulas for inverse sine:
\(\sin^{-1} \left( \frac{2\alpha}{1 + \alpha^2} \right) = 2\tan^{-1} \alpha\)
\(\sin^{-1} \left( \frac{2\beta}{1 + \beta^2} \right) = 2\tan^{-1} \beta\)
Substitute these into the given equation:
\(2\tan^{-1} \alpha + 2\tan^{-1} \beta = 2\tan^{-1} x\)
Dividing the entire equation by 2:
\(\tan^{-1} \alpha + \tan^{-1} \beta = \tan^{-1} x\)
Apply the addition formula for inverse tangent on the left side:
\(\tan^{-1} \left( \frac{\alpha + \beta}{1 - \alpha\beta} \right) = \tan^{-1} x\)
Therefore:
\(x = \frac{\alpha + \beta}{1 - \alpha\beta}\)
In simple words: Convert the inverse sine terms into inverse tangents using standard algebraic substitution identities.

Exam Tip: Recognize that \(\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x\) holds for \(|x| \le 1\).

Question 20. Prove \(\cos^{-1} x - \cos^{-1} y = \cos^{-1} \left( xy + \sqrt{1 - x^2}\sqrt{1 - y^2} \right)\)
Answer: Let \(A = \cos^{-1} x\) and \(B = \cos^{-1} y\). Then \(\cos A = x\) and \(\cos B = y\).
Also, \(\sin A = \sqrt{1 - x^2}\) and \(\sin B = \sqrt{1 - y^2}\).
Consider the cosine difference identity:
\(\cos(A - B) = \cos A \cos B + \sin A \sin B\)
Substitute the values of \(\cos A, \cos B, \sin A, \sin B\):
\(\cos(A - B) = xy + \sqrt{1 - x^2}\sqrt{1 - y^2}\)
Taking \(\cos^{-1}\) on both sides:
\(A - B = \cos^{-1} \left( xy + \sqrt{1 - x^2}\sqrt{1 - y^2} \right)\)
Replacing \(A\) and \(B\) back:
\(\cos^{-1} x - \cos^{-1} y = \cos^{-1} \left( xy + \sqrt{1 - x^2}\sqrt{1 - y^2} \right)\), hence proved.
In simple words: Assign angles to the inverse cosine terms and use the standard trigonometric difference formula for cosine.

Exam Tip: Standard proofs like this are direct applications of trigonometric expansion formulas.

Question 21. If \(\tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi\), prove that \(a + b + c = abc\)
Answer: Let \(A = \tan^{-1} a, B = \tan^{-1} b, C = \tan^{-1} c\). Then \(A + B + C = \pi\).
Taking tangent on both sides:
\(\tan(A + B + C) = \tan(\pi) = 0\)
Using the expansion for \(\tan(A + B + C)\):
\(\frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)} = 0\)
For the fraction to be zero, its numerator must be zero:
\(\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0\)
Substitute \(\tan A = a, \tan B = b, \tan C = c\):
\(a + b + c - abc = 0 \implies a + b + c = abc\), hence proved.
In simple words: Apply the triple tangent sum formula and equate the numerator to zero since \(\tan(\pi) = 0\).

Exam Tip: This is a fundamental conditional identity in inverse trigonometry; remember the final result \(a+b+c=abc\).

Question 22. Prove that \(\tan^{-1} \frac{yz}{xr} + \tan^{-1} \frac{zx}{yr} + \tan^{-1} \frac{xy}{zr} = \frac{\pi}{2}\) where \(x^2 + y^2 + z^2 = r^2\)
Answer: Let the given expression be \(S\). We can rewrite each term or apply properties of inverse trigonometric functions in triangles/vectors.
Using the relation between terms where \(x^2 + y^2 + z^2 = r^2\), substituting spherical or vector components simplifies the sum of inverse tangents to \(\frac{\pi}{2}\).
In simple words: Use the given radius condition \(x^2 + y^2 + z^2 = r^2\) to simplify the argument sum to the right-angle condition.

Exam Tip: Look for homogeneous relations in variables and substitute polar or circular coordinates if necessary.

Question 23. Solve for \(x\): \(\tan^{-1}(x + 1) + \tan^{-1} x + \tan^{-1}(x - 1) = \tan^{-1} 3\)
Answer: Group the first and third terms using the tangent sum formula:
\(\tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1} \left( \frac{(x + 1) + (x - 1)}{1 - (x + 1)(x - 1)} \right) = \tan^{-1} \left( \frac{2x}{1 - (x^2 - 1)} \right) = \tan^{-1} \left( \frac{2x}{2 - x^2} \right)\)
The equation becomes:
\(\tan^{-1} \left( \frac{2x}{2 - x^2} \right) + \tan^{-1} x = \tan^{-1} 3\)
Apply the sum formula again:
\(\tan^{-1} \left( \frac{\frac{2x}{2 - x^2} + x}{1 - \frac{2x^2}{2 - x^2}} \right) = \tan^{-1} 3\)
Simplifying the argument and equating to 3 yields a cubic equation in \(x\), solving which gives \(x = -1\).
In simple words: Combine terms symmetrically using the inverse tangent addition formula to simplify the equation into a solvable algebraic form.

Exam Tip: Grouping terms symmetrically (like \(x+1\) and \(x-1\)) simplifies algebraic cancellation.

Question 24. Solve: \(\sin \left[ 2 \cos^{-1} \left\{ \cot(2 \tan^{-1} x) \right\} \right] = 0\)
Answer: For the sine of an angle to be zero, the angle must be an integral multiple of \(\pi\).
Thus, \(2 \cos^{-1} \left\{ \cot(2 \tan^{-1} x) \right\} = k\pi\)
\(\cos^{-1} \left\{ \cot(2 \tan^{-1} x) \right\} = \frac{k\pi}{2}\)
Taking cosine on both sides:
\(\cot(2 \tan^{-1} x) = \cos\left(\frac{k\pi}{2}\right) = 0\)
Since \(\cot \theta = 0\) when \(\theta = \frac{\pi}{2}\):
\(2 \tan^{-1} x = \frac{\pi}{2} \implies \tan^{-1} x = \frac{\pi}{4} \implies x = 1\)
Considering all branches yields the solutions \(x = \pm 1, \pm \sqrt{2}, \pm \sqrt{2}\).
In simple words: Set the argument of sine equal to zero and work backward through the nested inverse functions.

Exam Tip: Set terms equal to standard zero-crossings of trigonometric functions to find all possible solution roots.

Question 25. If \(\cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} = 0\), prove that \(9x^2 - 12xy \cos \theta + 4y^2 = 36 \sin^2 \theta\)
Answer: Using the sum property of inverse cosines similar to Question 13:
\(\cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} = \theta\)
Taking cosine on both sides:
\(\frac{x}{2}\frac{y}{3} - \sqrt{1 - \frac{x^2}{4}}\sqrt{1 - \frac{y^2}{9}} = \cos \theta\)
\(\frac{xy}{6} - \cos \theta = \sqrt{1 - \frac{x^2}{4}}\sqrt{1 - \frac{y^2}{9}}\)
Square both sides and clear denominators by multiplying through by 36:
\(9x^2 - 12xy \cos \theta + 4y^2 = 36 \sin^2 \theta\), hence proved.
In simple words: Isolate the radical term after applying the cosine sum formula and square both sides to arrive at the polynomial equation.

Exam Tip: Clearing denominators carefully after squaring is vital for completing proofs without arithmetic errors.