🔥 Super Revision Test – Inverse Trigonometric Functions (2026)
CBSE Class 12 (2026) — 101 carefully curated questions with solutions.
📋 Questions
- Find the principal value of \( \sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right) \).
- Evaluate \( \cos^{-1}\!\left(-\dfrac{1}{2}\right) \).
- Find \( \tan^{-1}(1) + \tan^{-1}(2) \).
- Compute \( \sin^{-1}\!\left(\dfrac{3}{5}\right) + \cos^{-1}\!\left(\dfrac{4}{5}\right) \).
- Find \( \tan^{-1}\!\left(\dfrac{1}{3}\right) + \tan^{-1}\!\left(\dfrac{1}{2}\right) \).
- Show value of \( \sin^{-1}x + \cos^{-1}x \) for \( -1\le x\le1 \).
- Evaluate \( \tan^{-1}(2) - \tan^{-1}(3) \).
- Compute \( \sin^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) + \cos^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \).
- Find principal value of \( \cot^{-1}(\sqrt{3}) \).
- Find \( \sec^{-1}(2) \) (principal value).
- Evaluate \( \tan^{-1}\!\left(\dfrac{1}{7}\right) + \tan^{-1}\!\left(\dfrac{1}{3}\right) \).
- Find \( \sin^{-1}x + \sin^{-1}\!\left(\sqrt{1-x^2}\right) \) for \(0\le x\le1\).
- Find principal value \( \tan^{-1}(-1) \).
- Evaluate \( \sin^{-1}\!\left(-\dfrac{1}{2}\right) \).
- Find \( \cos^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \).
- Compute \( \tan^{-1}(2) + \tan^{-1}(3) \).
- Find \( \sin^{-1}\!\left(\dfrac{1}{3}\right) + \cos^{-1}\!\left(\dfrac{\sqrt{8}}{3}\right) \) if defined.
- Find \( \tan^{-1}x + \tan^{-1}\!\left(\dfrac{1}{x}\right) \) for \(x>0\).
- Find principal value of \( \sin^{-1}\!\left(\dfrac{24}{25}\right) \).
- Evaluate \( \cos^{-1}\!\left(-\dfrac{4}{5}\right) \).
- Verify: \( \sin^{-1}x + \sin^{-1}y = \sin^{-1}\!\big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\big) \) under suitable conditions.
- If \( \tan^{-1}a + \tan^{-1}b = \dfrac{\pi}{4} \), find relation between \(a\) and \(b\).
- Find the range of \( y=\tan^{-1}(\sin x) \).
- Find domain of \( f(x)=\cos^{-1}\!\big(\dfrac{2x}{1+x^2}\big) \).
- Evaluate \( \sin(\sin^{-1}x) \) for \( -1\le x\le1 \).
- Find \( \cos(\tan^{-1}x) \) in terms of \(x\).
- Compute \( \tan(\cos^{-1}x) \) for \( -1\le x\le1 \), \(x\ne0\).
- Find \( \sin^{-1}\!\left(\dfrac{12}{13}\right) + \cos^{-1}\!\left(\dfrac{5}{13}\right) \).
- Find \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) \).
- Find principal value of \( \cot^{-1}(-1) \).
- Evaluate \( \tan^{-1}\!\left(\dfrac{1}{2}\right) + \tan^{-1}\!\left(\dfrac{1}{5}\right) + \tan^{-1}\!\left(\dfrac{1}{8}\right) \).
- Solve for \(x\): \( \sin^{-1}\!x = \dfrac{\pi}{6} \).
- Simplify \( \tan^{-1}\!\left(\dfrac{3x-x^3}{1-3x^2}\right) \) (identify relation to \( \tan^{-1}x \)).
- Express \( \sin^{-1}\!\left(\dfrac{1-2x^2}{1+2x^2}\right) \) in terms of \( \tan^{-1}x \) if possible.
- Find value of \( \tan^{-1}x + \tan^{-1}\!\left(\dfrac{1-x}{1+x}\right) \) (state valid \(x\)).
- If \( \sin^{-1}x + \sin^{-1}y = \dfrac{\pi}{3} \), express \(y\) in terms of \(x\).
- Evaluate \( \cos^{-1}\!\left(\dfrac{3}{5}\right) - \sin^{-1}\!\left(\dfrac{4}{5}\right) \).
- Find smallest positive \(x\) such that \( \tan^{-1}x = \sin^{-1}\!\left(\dfrac{3x}{1+x^2}\right) \).
- Express \( \tan^{-1}\!\left(\dfrac{2ab}{1-a^2b^2}\right) \) in terms of \( \tan^{-1}a \) and \( \tan^{-1}b \) with conditions.
- If \( \tan^{-1}x + \tan^{-1}y = \dfrac{\pi}{4} \), find \(y\) in terms of \(x\).
- Prove \( \sin^{-1}x + \sin^{-1}y = \cos^{-1}\!\big(\sqrt{1-x^2}\sqrt{1-y^2} - xy\big) \) under domain conditions.
- Find \( \tan^{-1}\!\left(\dfrac{3}{4}\right) + \tan^{-1}\!\left(\dfrac{4}{3}\right) \).
- Range of \( f(x)=\sin^{-1}\!\big(\dfrac{2x}{1+x^2}\big) \) for real \(x\).
- Check whether \( \tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi \) (principal-value comment).
- Find \( \tan^{-1}x - \tan^{-1}\!\left(\dfrac{1}{x}\right) \) when \( x<0 li=""> 0>
- Find \( \sin^{-1}(\sin 2) \) (principal value in radians).
- Find \( \cos^{-1}(\cos 3) \) (principal value in radians).
- Differentiate \( y = \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right) \).
- Find \( \sin^{-1}x - \cos^{-1}x \) for \( -1\le x\le1 \).
- Compute \( \tan^{-1}(\sqrt{3}) + \tan^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right) \).
- Find principal value of \( \sec^{-1}(-2) \).
- Find \(x,y,z\in[-1,1]\) satisfying \( \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi \) in a simple special case.
- Show \( \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi \) if \( xy+yz+zx=1 \) under principal branches.
- Principal value of \( \sin^{-1}\!\left(-\dfrac{\sqrt{2}}{2}\right) \).
- Compute \( \tan^{-1}\!\left(\dfrac{5}{12}\right) + \tan^{-1}\!\left(\dfrac{12}{5}\right) \).
- Find \( \sin^{-1}\!\left(\dfrac{8}{17}\right) + \cos^{-1}\!\left(\dfrac{15}{17}\right) \).
- Evaluate \( \tan^{-1}\!\left(\dfrac{x^2-1}{2x}\right) \) relation to half-angle.
- Find \( \cos(\sin^{-1}x) \) in terms of \(x\).
- Find \( \sin(\cos^{-1}x) \) in terms of \(x\).
- Compute \( \tan^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right) + \tan^{-1}\!\left(\dfrac{2}{\sqrt{3}}\right) \).
- Principal value of \( \csc^{-1}(-1) \).
- Find \( x \) such that \( \sin^{-1}\!x = \cos^{-1}\!\left(\dfrac{1}{2}\right) \).
- Solve \( \tan^{-1}\!x + \tan^{-1}\!2x = \dfrac{\pi}{4} \) for real \(x\).
- Show \( \tan(\sin^{-1}x) = \dfrac{x}{\sqrt{1-x^2}} \).
- Principal value \( \tan^{-1}\!\left(-\dfrac{1}{\sqrt{3}}\right) \).
- Find \( \sin^{-1}\!\left(\dfrac{4}{5}\right) - \cos^{-1}\!\left(\dfrac{3}{5}\right) \).
- Evaluate \( \tan^{-1}\!\left(\dfrac{7}{24}\right) + \tan^{-1}\!\left(\dfrac{24}{7}\right) \).
- Find \( \sin^{-1}\!\left(\dfrac{20}{29}\right) + \cos^{-1}\!\left(\dfrac{21}{29}\right) \).
- Prove \( \tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right) \) with principal-value note.
- Find \( \sin^{-1}\!\left(\dfrac{2}{\sqrt{5}}\right) + \cos^{-1}\!\left(\dfrac{1}{\sqrt{5}}\right) \).
- Evaluate \( \tan^{-1}\!\left(\dfrac{9}{40}\right) + \tan^{-1}\!\left(\dfrac{40}{9}\right) \).
- Solve \( \sin^{-1}x + \sin^{-1}\!\left(\dfrac{3}{5}\right) = \dfrac{\pi}{2} \) for \(x\).
- State domain where \( \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right) = 2\tan^{-1}x \) holds.
- Compute \( \sin^{-1}0 + \cos^{-1}0 \).
- Simplify \( \tan^{-1}x + \tan^{-1}\!\left(\dfrac{1-x}{1+x}\right) \) for \(x\ne -1\).
- Principal value \( \cos^{-1}(-1) \).
- Find whether \( \sin^{-1}\!\left(\dfrac{3}{5}\right) + \sin^{-1}\!\left(\dfrac{4}{5}\right) \) is valid; if so compute.
- Show \( \sin^{-1}\!\left(\dfrac{2t}{1+t^2}\right) = 2\tan^{-1}t \) under suitable \(t\).
- Compute \( \tan^{-1}1 + \cot^{-1}1 \) (principal values).
- Compute \( \sin^{-1}\!\left(\dfrac{5}{13}\right) + \cos^{-1}\!\left(\dfrac{12}{13}\right) \).
- Principal value of \( \csc^{-1}(1) \).
- Express \( \tan^{-1}\!\left(\dfrac{4x}{1-x^2}\right) \) in terms of \( \tan^{-1}x \) if possible.
- Solve \( \tan^{-1}x + \tan^{-1}y = 0 \) for relation between \(x\) and \(y\).
- Find \( \sin^{-1}\!\left(\dfrac{7}{25}\right) + \cos^{-1}\!\left(\dfrac{24}{25}\right) \).
- Find derivative of \( y=\sin^{-1}x \).
- Prove \( \tan^{-1}\!\left(\dfrac{a-b}{1+ab}\right) = \tan^{-1}a - \tan^{-1}b \) with principal-value remark.
- Find \( \sin^{-1}\!\left(\dfrac{15}{17}\right) + \cos^{-1}\!\left(\dfrac{8}{17}\right) \).
- Evaluate \( \tan^{-1}\!\left(\dfrac{1}{2}\right) + \tan^{-1}\!\left(\dfrac{2}{3}\right) + \tan^{-1}\!\left(\dfrac{3}{4}\right) \).
- Show relation: \( \sec^{-1}(2)+\sec^{-1}(-2)=\pi \) (principal branch note).
- Given \( \sin2\theta=x \), express \( \sin^{-1}x \) in terms of \( \theta \) (choose simplest form).
- Evaluate \( \tan^{-1}\!\left(\dfrac{11}{60}\right) + \tan^{-1}\!\left(\dfrac{60}{11}\right) \).
- Relate \( \sin^{-1}x \) and \( \tan^{-1}y \) if \( \sin^{-1}x = \tan^{-1}y \).
- Compute \( \sin^{-1}\!\left(\dfrac{\sqrt{5}-1}{4}\right) + \sin^{-1}\!\left(\dfrac{\sqrt{5}+1}{4}\right) \) (special values).
- Find relation \( \tan^{-1}\!\left(\dfrac{x}{1-x^2}\right) \) to double-angle form.
- Principal value \( \cos^{-1}\!\left(\dfrac{1}{2}\right) \).
- Find \( \sin^{-1}\!\left(\dfrac{1}{\sqrt{10}}\right) + \cos^{-1}\!\left(\dfrac{3}{\sqrt{10}}\right) \).
- Prove \( \sin^{-1}x = \tan^{-1}\!\big(\dfrac{x}{\sqrt{1-x^2}}\big) \) for \( -1
- Evaluate \( \tan^{-1}\!\left(\dfrac{2}{5}\right) + \tan^{-1}\!\left(\dfrac{5}{12}\right) + \tan^{-1}\!\left(\dfrac{12}{5}\right) \).
- Principal value of \( \cot^{-1}0 \).
- Range of \( y=\cos^{-1}\!\big(\dfrac{1-x^2}{1+x^2}\big) \) for real \(x\).
- Final: Show \( \tan^{-1}\!\left(\dfrac{3x-x^3}{1-3x^2}\right) = 3\tan^{-1}x \) under principal-value conditions (identify condition).
🧠 Short Solutions (1–2 lines each)
Click to View Answers
1. \( \sin^{-1}(\sqrt{3}/2)=\pi/3\) since \( \sin(\pi/3)=\sqrt{3}/2\).
2. \( \cos^{-1}(-1/2)=2\pi/3\) because \( \cos(2\pi/3)=-1/2\) in \([0,\pi]\).
3. \( \tan^{-1}1+\tan^{-1}2=\tan^{-1}\!\left(\dfrac{1+2}{1-2}\right)=\tan^{-1}(-3)\); add \(\pi\) for principal value → \(3\pi/4\).
4. \( \sin^{-1}(3/5)+\cos^{-1}(4/5)=\pi/2\) since they are complementary in a \(3\!-\!4\!-\!5\) triangle.
5. Use arctan addition: \( \tan^{-1}(1/3)+\tan^{-1}(1/2)=\tan^{-1}\!\left(\dfrac{(1/3)+(1/2)}{1-(1/6)}\right)=\tan^{-1}\!\left(\dfrac{5/6}{5/6}\right)=\tan^{-1}1=\pi/4.\)
6. \( \sin^{-1}x+\cos^{-1}x=\pi/2\) for all \(x\in[-1,1]\) (principal branches).
7. \( \tan^{-1}2-\tan^{-1}3=\tan^{-1}\!\left(\dfrac{2-3}{1+6}\right)=\tan^{-1}(-1/7)=-\tan^{-1}(1/7).\)
8. \( \sin^{-1}(1/\sqrt2)+\cos^{-1}(1/\sqrt2)=\pi/2\) (both correspond to \(\pi/4\) complementary).
9. \( \cot^{-1}(\sqrt3)=\pi/6\) in the principal range \((0,\pi)\) since \(\cot(\pi/6)=\sqrt3.\)
10. \( \sec^{-1}(2)=\pi/3\) because \(\sec(\pi/3)=2\) and principal value in \([0,\pi],\; \ne \pi/2\).
11. Use formula: \(\tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)\). For \(1/7,1/3\): sum = \(\tan^{-1}(10/21)\) (principal numeric value).
12. For \(0\le x\le1\), \(\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}=\pi/2\) since \(\sqrt{1-x^2}=\cos(\sin^{-1}x)\).
13. \( \tan^{-1}(-1)=-\pi/4\) in \((-\pi/2,\pi/2)\).
14. \( \sin^{-1}(-1/2)=-\pi/6\) (principal arcsin range \([- \pi/2,\pi/2]\)).
15. \( \cos^{-1}(1/\sqrt2)=\pi/4\) as \(\cos(\pi/4)=1/\sqrt2\).
16. \(\tan^{-1}2+\tan^{-1}3=\tan^{-1}\!\left(\dfrac{5}{-5}\right)+\pi=\tan^{-1}(-1)+\pi=\pi-\pi/4=3\pi/4\) (branch correction).
17. Compute: \(\sin^{-1}(1/3)+\cos^{-1}(\sqrt8/3)\). Note \((1/3)^2+( \sqrt8/3)^2=1\) → sum \(=\pi/2\).
18. For \(x>0\), \(\tan^{-1}x+\tan^{-1}(1/x)=\pi/2\) (since reciprocals complement in principal range).
19. Leave as arcsin: principal value \( \sin^{-1}(24/25)\) (approx \( \approx 1.3181\) rad) — corresponds to a \(7,24,25\) triangle.
20. \(\cos^{-1}(-4/5)=\pi-\cos^{-1}(4/5)\) (principal value in \([0,\pi]\)).
21. The addition formula holds when expressions inside arcsin produce value in \([-1,1]\) and sign choices match principal branches — derivable using sin(A+B).\)
22. From \(\tan(A+B)=1\): \(\dfrac{a+b}{1-ab}=1\) → \(a+b=1-ab\).\)
23. Range of \(\tan^{-1}(\sin x)\) is \([-\pi/4,\pi/4]\) because \(\sin x\in[-1,1]\) and \(\tan^{-1}\) of \(\pm1\) is \(\pm\pi/4\).
24. Domain is all real \(x\) because \(\dfrac{2x}{1+x^2}\in[-1,1]\) for all real \(x\) (AM-GM / rearrange inequality).
25. \(\sin(\sin^{-1}x)=x\) for \(x\in[-1,1]\) (definition of arcsin).
26. \(\cos(\tan^{-1}x)=\dfrac{1}{\sqrt{1+x^2}}\) using a right triangle with opposite \(x\) and adjacent \(1\).
27. \(\tan(\cos^{-1}x)=\dfrac{\sqrt{1-x^2}}{x}\) for \(x\ne0\) (triangle with adjacent \(x\)).
28. \( \sin^{-1}(12/13)+\cos^{-1}(5/13)=\pi/2\) (12-5-13 triangle: complementary angles).
29. Using earlier result, \( \tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi\) with principal branch corrections (standard identity).
30. \( \cot^{-1}(-1)=3\pi/4\) in principal range \((0,\pi)\) since \(\cot(3\pi/4)=-1.\)
31. Using triple reciprocal identities gives numeric simplification ≈ \(\pi/4\) (compute via pairwise additions → \(\pi/4\)).
32. \( \sin^{-1}x=\pi/6 \Rightarrow x=\sin(\pi/6)=1/2.\)
33. Use triple-angle arctan identity: \(\tan(3\theta)=\dfrac{3t-t^3}{1-3t^2}\) with \(t=\tan\theta\). So expression = \(3\tan^{-1}x\) if principal branch allowed.
34. Write \( (1-2x^2)/(1+2x^2)=\cos(2\phi) \) for \( \tan\phi=x\) or relate to \(2\tan^{-1}x\); simplifies to expression in terms of \(\tan^{-1}x\).
35. Use arctan addition: simplifies often to \(\pi/4\) for generic real \(x\) not equal to \(-1\); ensure branch check.
36. From \(\sin(A+B)=\sin(\pi/3)\) infer \(y=\sin(\pi/3-\sin^{-1}x)\); algebra leads to explicit expression \(y=\dfrac{\sqrt3\sqrt{1-x^2}-x}{2}\).
37. \(\cos^{-1}(3/5)-\sin^{-1}(4/5)=\cos^{-1}(3/5)-(\pi/2-\cos^{-1}(3/5))=2\cos^{-1}(3/5)-\pi/2\).
38. Solve equation — substitution yields \(x=\sqrt3-1\) as smallest positive root after algebraic simplification (check branch consistency).
39. \(\tan^{-1}\!\left(\dfrac{2ab}{1-a^2b^2}\right)=\tan^{-1}a+\tan^{-1}b\) if \(1-a^2b^2>0\); otherwise add \(\pi\) or \(-\pi\) for branch correction.
40. From addition formula get \(y=\dfrac{1-x}{1+x}\) when \(\tan(A+B)=1\) and principal branch used (algebraic rearrangement).
41. Derive via \(\cos(\frac{\pi}{2}-A-B)\) and identity manipulations—holds under principal branches (short proof expected in exam).
42. \(\tan^{-1}(3/4)+\tan^{-1}(4/3)=\tan^{-1}(\infty)=\pi/2\) (pair of reciprocals >0 sum to \(\pi/2\)).
43. Range of \(\sin^{-1}(2x/(1+x^2))\) is \([-\pi/2,\pi/2]\); actual values cover \((- \pi/2, \pi/2)\) with endpoints approached as \(x\to \pm\infty\).
44. Yes, with principal branch corrections the sum equals \(\pi\); demonstrate by pairwise addition and branch management (standard exercise).
45. For \(x<0 appropriately="" div="" each="" maps="" negative="" pi="" quadrant="" since="" tan="" to="" x-="" x="">
46. \(\sin^{-1}(\sin2)=\pi-2\) because \(2\) rad is in \((\pi/2,\pi)\) and principal arcsin maps to \(\pi-2\).
47. \(\cos^{-1}(\cos3)=\pi-3\) as \(3\) lies in \((\pi/2,\pi)\) so principal arccos returns \(\pi-3\).
48. Differentiate: \( y'=\dfrac{1}{1+\left(\frac{2x}{1-x^2}\right)^2}\cdot \dfrac{2(1-x^2)+2x(2x)}{(1-x^2)^2}\). Simplify to final rational form.
49. Using (6): \(\sin^{-1}x-\cos^{-1}x = 2\sin^{-1}x - \pi/2.\)
50. \(\tan^{-1}\sqrt3+\tan^{-1}(1/\sqrt3)=\pi/2\) (since product is 1 and both positive).
51. \(\sec^{-1}(-2)=\pi-\sec^{-1}(2)=\pi-\pi/3=2\pi/3\) in principal branch convention.\)
52. Example: take \(x=1, y=1, z=-1/2\) or construct from complementary angles producing sum \(\pi\) — e.g., \(x=\sin(\pi/3), y=\sin(\pi/6), z=\sin(\pi/2)\) check ranges; many special triples exist (show one consistent triple).
53. Using tangent sum identities and branch constraints, if \(xy+yz+zx=1\) and each arctan in principal range, sum \(\pi\) follows (sketch: set \(x=\tan A\) etc.).
54. \(\sin^{-1}(-\sqrt2/2)=-\pi/4\) (principal arcsin gives negative angle).
55. \(\tan^{-1}(5/12)+\tan^{-1}(12/5)=\pi/2\) (reciprocal pair positive → \(\pi/2\)).
56. \( \sin^{-1}(8/17)+\cos^{-1}(15/17)=\pi/2\) (8-15-17 right triangle complementarity).
57. \(\tan^{-1}((x^2-1)/(2x))\) relates to half-angle: set \(t=\tan^{-1}x\) and use formula for \(\tan(2t)\) or \(\tan(\pi/2-2t)\) to identify relation.
58. \(\cos(\sin^{-1}x)=\sqrt{1-x^2}\) (principal positive root for \(-1\le x\le1\)).
59. \(\sin(\cos^{-1}x)=\sqrt{1-x^2}\) (similar triangle reasoning).
60. Compute: \(\tan^{-1}(1/\sqrt3)=\pi/6, \tan^{-1}(2/\sqrt3)=\pi/3\) → sum = \(\pi/2\).
61. \(\csc^{-1}(-1)=-\pi/2\) (principal value corresponding to sine inverse −1 at \(-\pi/2\)).
62. Solve: \(\cos^{-1}(1/2)=\pi/3\). So \(\sin^{-1}x=\pi/3\Rightarrow x=\sin(\pi/3)=\sqrt3/2.\)
63. Solve \(\tan^{-1}x+\tan^{-1}2x=\pi/4\) using tangent addition: \(\dfrac{x+2x}{1-2x^2}=1 \Rightarrow 3x=1-2x^2\Rightarrow2x^2+3x-1=0\); solve → \(x=\dfrac{-3+\sqrt{17}}{4}\) (positive root) or the other root if acceptable by domain.
64. \(\tan(\sin^{-1}x)=\dfrac{x}{\sqrt{1-x^2}}\) (opposite/adjacent using triangle with hypotenuse 1).
65. \(\tan^{-1}(-1/\sqrt3)=-\pi/6\) (principal range).\)
66. Using complementary relations: \(\sin^{-1}(4/5)-\cos^{-1}(3/5)=\sin^{-1}(4/5)-(\pi/2-\sin^{-1}(3/5))\) → compute accordingly (gives \(2\sin^{-1}(4/5)-\pi/2\)).
67. \(\tan^{-1}(7/24)+\tan^{-1}(24/7)=\pi/2\) (positive reciprocals pair).
68. \(\sin^{-1}(20/29)+\cos^{-1}(21/29)=\pi/2\) (20-21-29 right triangle complementarity).
69. Use tangent addition formula; principal-value caution: requires \(1-xy\ne0\) and branch adjustments as needed (standard derivation).
70. \(\sin^{-1}(2/\sqrt5)+\cos^{-1}(1/\sqrt5)=\pi/2\) (complementary legs of right triangle with hypotenuse \(\sqrt5\)).
71. \(\tan^{-1}(9/40)+\tan^{-1}(40/9)=\pi/2\) (reciprocal positive pair → \(\pi/2\)).
72. Solve: \(\sin^{-1}x+\sin^{-1}(3/5)=\pi/2\Rightarrow \sin^{-1}x=\pi/2-\sin^{-1}(3/5)=\cos^{-1}(3/5)\Rightarrow x=\cos(\cos^{-1}(3/5))=4/5.\)
73. Identity \( \tan^{-1}(2x/(1-x^2))=2\tan^{-1}x \) holds when \(|x|<1 and="" branch="" condition="" div="" jump="" no="" occurs="" principal-range="">
0>74. \(\sin^{-1}0+\cos^{-1}0=0+\pi/2=\pi/2.\)
75. Simplify algebraically; typical result equals \(\pi/4\) for many real \(x\) (perform addition formula and simplify carefully with domain checks).
76. \(\cos^{-1}(-1)=\pi\) (principal arccos range [0,π]).
77. \(\sin^{-1}(3/5)+\sin^{-1}(4/5)\) is not valid because \(3/5\) and \(4/5\) correspond to complementary angles whose sum would be \(\pi/2\) only if both arcsin are within domain; check numeric: \(\sin^{-1}(3/5)\approx0.6435,\sin^{-1}(4/5)\approx0.9273\) sum ≈1.5708=\(\pi/2\) — it is valid and equals \(\pi/2\).
78. Using substitution \(t=\tan^{-1}t\): \(\sin^{-1}(2t/(1+t^2))=2\tan^{-1}t\) with domain restrictions on \(t\) (same identity as earlier).
79. \(\tan^{-1}1+\cot^{-1}1=\pi/2\) using principal values (\(\tan^{-1}1=\pi/4\), \(\cot^{-1}1=\pi/4\)).
80. \(\sin^{-1}(5/13)+\cos^{-1}(12/13)=\pi/2\) (5-12-13 triangle complementarity).
81. \(\csc^{-1}(1)=\pi/2\) principal value (since \(\csc(\pi/2)=1\)).
82. Use double-angle arctan formula: \(\tan^{-1}(4x/(1-x^2))=2\tan^{-1}(2x/(1-x^2))\) with branch caveat; express in terms of \(\tan^{-1}x\) with algebraic steps.
83. \(\tan^{-1}x+\tan^{-1}y=0\Rightarrow y=-x\) provided both arctans are in principal domain (simple relation).
84. \(\sin^{-1}(7/25)+\cos^{-1}(24/25)=\pi/2\) (7-24-25 triangle-like complement; if exact triangle holds sum=π/2).
85. \( \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}\) for \(|x|<1 div="">
1>86. Derive via tangent subtraction formula; principal-value remark: if \(1+ab\ne0\) then identity holds, else adjust by \(\pi\).
87. \(\sin^{-1}(15/17)+\cos^{-1}(8/17)=\pi/2\) (15-8-17 triangle complementarity).
88. Numeric evaluation yields \(\dfrac{3\pi}{4}\) for the three-term arctan sum in many standard patterns (explicit compute by pairwise addition and branch check).
89. \(\sec^{-1}(2)+\sec^{-1}(-2)=\pi\) because they are supplements in principal branches.\)
90. If \(\sin2\theta=x\) then \(\sin^{-1}x=2\theta\) or \( \pi-2\theta\) depending on range; simplest: \( \sin^{-1}x=2\theta\) if \(2\theta\in[-\pi/2,\pi/2]\).
91. \(\tan^{-1}(11/60)+\tan^{-1}(60/11)=\pi/2\) (positive reciprocal pair → \(\pi/2\)).
92. If \(\sin^{-1}x=\tan^{-1}y\) then \(x=\dfrac{y}{\sqrt{1+y^2}}\) (convert \(y=\tan\alpha\Rightarrow x=\sin\alpha\)).
93. The special golden-ratio-like values sum to \(\pi/2\) (compute numeric or use trig identities to show sum = π/2).
94. Express \(\tan^{-1}\!\left(\dfrac{x}{1-x^2}\right)\) using double-angle relations for \(\tan^{-1}\); relates to \(\tan^{-1}x\) and \(\tan^{-1}(-x)\) depending on domain — manipulate via sum formula.
95. \(\cos^{-1}(1/2)=\pi/3\) (standard value).
96. \(\sin^{-1}(1/\sqrt{10})+\cos^{-1}(3/\sqrt{10})=\pi/2\) if \( (1^2+3^2)/10=1\) i.e., they form complementary legs — check numerically: yes → \(\pi/2\).
97. Starting from \(x=\sin\theta\) get \(\sin^{-1}x=\tan^{-1}\!\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) using triangle with opposite \(x\), adjacent \(\sqrt{1-x^2}\).
98. Evaluate triple arctan: pairwise reduce → typical result \(\dfrac{3\pi}{4}\) depending on values; do pairwise additions and branch checks.
99. \(\cot^{-1}0=\pi/2\) in principal branch (since \(\cot(\pi/2)=0\)).
100. Range of \( \cos^{-1}\!\big(\dfrac{1-x^2}{1+x^2}\big)\) : since argument ranges \([-1,1]\), arccos range is \([0,\pi]\); mapping of real \(x\) covers \([0,\pi]\) with specifics: as \(x\) varies from \(-\infty\) to \(\infty\), value goes from 0 to \(\pi\).
101. Triple-angle identity: \(\tan^{-1}\!\left(\dfrac{3x-x^3}{1-3x^2}\right)=3\tan^{-1}x\) provided \(x\) chosen so that \(3\tan^{-1}x\) lies in \((-\pi/2,\pi/2)\); otherwise add/subtract \(\pi\) as branch correction.
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