This MCQ module is based on: Exercises and Summary – Inverse Trigonometric Functions
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Exercises and Summary – Inverse Trigonometric Functions
🎓 Class 12
Mathematics
CBSE
Theory
Ch 2 — Inverse Trigonometric Functions
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Exercises and Summary – Inverse Trigonometric Functions
Targeting Class 12 level in Trigonometry, with Advanced difficulty.
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Miscellaneous Examples
Example 6
Find the value of \(\sin^{-1}\left(\sin\frac{3\pi}{5}\right)\).
Solution
We know that \(\sin^{-1}(\sin x) = x\) only if \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Since \(\frac{3\pi}{5} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we cannot directly simplify.We write: \(\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}\).
Now \(\frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) (since \(\frac{2\pi}{5} = 0.4\pi < \frac{\pi}{2}\)).
Therefore, \(\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\frac{2\pi}{5}\right) = \frac{2\pi}{5}\).
Common Mistake Alert
Students often write \(\sin^{-1}(\sin x) = x\) for all \(x\). This is wrong! The cancellation \(\sin^{-1}(\sin x) = x\) holds ONLY when \(x\) lies in the principal value range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). If \(x\) is outside this range, you must first convert \(\sin x\) to \(\sin y\) where \(y\) is in the principal range.
Miscellaneous Exercise on Chapter 2
Find the value of the following:
Q1. \(\cos^{-1}\left(\cos\frac{13\pi}{6}\right)\)
\(\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}\). So \(\cos\frac{13\pi}{6} = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\).
Since \(\frac{\pi}{6} \in [0, \pi]\): \(\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}\).
Since \(\frac{\pi}{6} \in [0, \pi]\): \(\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}\).
Q2. \(\tan^{-1}\left(\tan\frac{7\pi}{6}\right)\)
\(\frac{7\pi}{6} = \pi + \frac{\pi}{6}\). So \(\tan\frac{7\pi}{6} = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\).
Since \(\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\): \(\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
Since \(\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\): \(\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
Prove that:
Q3. \(2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}\)
Let \(\sin^{-1}\frac{3}{5} = \theta\). Then \(\sin\theta = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\), \(\tan\theta = \frac{3}{4}\).
\(\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}\).
So \(2\theta = \tan^{-1}\frac{24}{7}\), i.e., \(2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}\).
\(\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}\).
So \(2\theta = \tan^{-1}\frac{24}{7}\), i.e., \(2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}\).
Q4. \(\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36}\)
Let \(\alpha = \sin^{-1}\frac{8}{17}\) and \(\beta = \sin^{-1}\frac{3}{5}\).
Then: \(\sin\alpha = \frac{8}{17}, \cos\alpha = \frac{15}{17}, \tan\alpha = \frac{8}{15}\).
And: \(\sin\beta = \frac{3}{5}, \cos\beta = \frac{4}{5}, \tan\beta = \frac{3}{4}\).
\(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}} = \frac{\frac{32 + 45}{60}}{1 - \frac{24}{60}} = \frac{\frac{77}{60}}{\frac{36}{60}} = \frac{77}{36}\).
Hence \(\alpha + \beta = \tan^{-1}\frac{77}{36}\).
Then: \(\sin\alpha = \frac{8}{17}, \cos\alpha = \frac{15}{17}, \tan\alpha = \frac{8}{15}\).
And: \(\sin\beta = \frac{3}{5}, \cos\beta = \frac{4}{5}, \tan\beta = \frac{3}{4}\).
\(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}} = \frac{\frac{32 + 45}{60}}{1 - \frac{24}{60}} = \frac{\frac{77}{60}}{\frac{36}{60}} = \frac{77}{36}\).
Hence \(\alpha + \beta = \tan^{-1}\frac{77}{36}\).
Q5. \(\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}\)
Let \(\alpha = \cos^{-1}\frac{4}{5}\) and \(\beta = \cos^{-1}\frac{12}{13}\).
Then: \(\cos\alpha = \frac{4}{5}, \sin\alpha = \frac{3}{5}\) and \(\cos\beta = \frac{12}{13}, \sin\beta = \frac{5}{13}\).
\(\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}\).
Hence \(\alpha + \beta = \cos^{-1}\frac{33}{65}\).
Then: \(\cos\alpha = \frac{4}{5}, \sin\alpha = \frac{3}{5}\) and \(\cos\beta = \frac{12}{13}, \sin\beta = \frac{5}{13}\).
\(\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}\).
Hence \(\alpha + \beta = \cos^{-1}\frac{33}{65}\).
Q6. \(\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}\)
Let \(\alpha = \cos^{-1}\frac{12}{13}\) so \(\cos\alpha = \frac{12}{13}, \sin\alpha = \frac{5}{13}\).
Let \(\beta = \sin^{-1}\frac{3}{5}\) so \(\sin\beta = \frac{3}{5}, \cos\beta = \frac{4}{5}\).
\(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}\).
Hence \(\alpha + \beta = \sin^{-1}\frac{56}{65}\).
Let \(\beta = \sin^{-1}\frac{3}{5}\) so \(\sin\beta = \frac{3}{5}, \cos\beta = \frac{4}{5}\).
\(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}\).
Hence \(\alpha + \beta = \sin^{-1}\frac{56}{65}\).
Q7. \(\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}\)
Let \(\alpha = \sin^{-1}\frac{5}{13}\), so \(\tan\alpha = \frac{5}{12}\).
Let \(\beta = \cos^{-1}\frac{3}{5}\), so \(\tan\beta = \frac{4}{3}\).
\(\tan(\alpha + \beta) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \cdot \frac{4}{3}} = \frac{\frac{5 + 16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{21}{12} \times \frac{36}{16} = \frac{756}{192} = \frac{63}{16}\).
Hence \(\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1}\frac{63}{16}\).
Let \(\beta = \cos^{-1}\frac{3}{5}\), so \(\tan\beta = \frac{4}{3}\).
\(\tan(\alpha + \beta) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \cdot \frac{4}{3}} = \frac{\frac{5 + 16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{21}{12} \times \frac{36}{16} = \frac{756}{192} = \frac{63}{16}\).
Hence \(\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1}\frac{63}{16}\).
Prove that:
Q8. \(\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)\), \(x \in [0, 1]\)
Let \(\sqrt{x} = \tan\theta\), so \(x = \tan^2\theta\) and \(\theta = \tan^{-1}\sqrt{x}\) where \(\theta \in \left[0, \frac{\pi}{4}\right]\).
RHS = \(\frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \frac{1}{2}\cos^{-1}(\cos 2\theta) = \frac{1}{2} \cdot 2\theta = \theta\) = LHS.
(Using the identity \(\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}\) and noting \(2\theta \in [0, \frac{\pi}{2}] \subset [0, \pi]\).)
RHS = \(\frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \frac{1}{2}\cos^{-1}(\cos 2\theta) = \frac{1}{2} \cdot 2\theta = \theta\) = LHS.
(Using the identity \(\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}\) and noting \(2\theta \in [0, \frac{\pi}{2}] \subset [0, \pi]\).)
Q9. \(\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}\), \(x \in \left(0, \frac{\pi}{4}\right)\)
Using the identities: \(\sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}\) and \(\sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}\) (for \(x \in (0, \frac{\pi}{2})\)):
Numerator = \((\cos\frac{x}{2} + \sin\frac{x}{2}) + (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\cos\frac{x}{2}\).
Denominator = \((\cos\frac{x}{2} + \sin\frac{x}{2}) - (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\sin\frac{x}{2}\).
The expression inside \(\cot^{-1}\) becomes \(\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}\).
So LHS = \(\cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}\) (since \(\frac{x}{2} \in (0, \frac{\pi}{8}) \subset (0, \pi)\)).
Numerator = \((\cos\frac{x}{2} + \sin\frac{x}{2}) + (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\cos\frac{x}{2}\).
Denominator = \((\cos\frac{x}{2} + \sin\frac{x}{2}) - (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\sin\frac{x}{2}\).
The expression inside \(\cot^{-1}\) becomes \(\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}\).
So LHS = \(\cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}\) (since \(\frac{x}{2} \in (0, \frac{\pi}{8}) \subset (0, \pi)\)).
Solve the following equations:
Q11. \(2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)\)
Using \(2\tan^{-1} t = \tan^{-1}\frac{2t}{1-t^2}\):
LHS = \(\tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right)\).
RHS = \(\tan^{-1}\left(\frac{2}{\sin x}\right)\).
Equating: \(\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}\), giving \(\frac{\cos x}{\sin x} = 1\), so \(\tan x = 1\).
\(x = \frac{\pi}{4}\) (principal value).
LHS = \(\tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right)\).
RHS = \(\tan^{-1}\left(\frac{2}{\sin x}\right)\).
Equating: \(\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}\), giving \(\frac{\cos x}{\sin x} = 1\), so \(\tan x = 1\).
\(x = \frac{\pi}{4}\) (principal value).
Q12. \(\tan^{-1}\frac{1-x}{1+x} = \frac{1}{2}\tan^{-1} x\), (\(x > 0\))
Note that \(\frac{1-x}{1+x} = \tan\left(\frac{\pi}{4} - \tan^{-1} x\right)\) (using \(\tan(A - B)\) with \(A = \frac{\pi}{4}\)).
So \(\tan^{-1}\frac{1-x}{1+x} = \frac{\pi}{4} - \tan^{-1} x\).
Given equation: \(\frac{\pi}{4} - \tan^{-1} x = \frac{1}{2}\tan^{-1} x\).
\(\frac{\pi}{4} = \frac{3}{2}\tan^{-1} x\), so \(\tan^{-1} x = \frac{\pi}{6}\), giving \(x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\).
So \(\tan^{-1}\frac{1-x}{1+x} = \frac{\pi}{4} - \tan^{-1} x\).
Given equation: \(\frac{\pi}{4} - \tan^{-1} x = \frac{1}{2}\tan^{-1} x\).
\(\frac{\pi}{4} = \frac{3}{2}\tan^{-1} x\), so \(\tan^{-1} x = \frac{\pi}{6}\), giving \(x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\).
Q13. \(\sin(\tan^{-1} x)\), \(|x| < 1\) is equal to
(A) \(\frac{x}{\sqrt{1-x^2}}\) (B) \(\frac{1}{\sqrt{1-x^2}}\) (C) \(\frac{1}{\sqrt{1+x^2}}\) (D) \(\frac{x}{\sqrt{1+x^2}}\)
(A) \(\frac{x}{\sqrt{1-x^2}}\) (B) \(\frac{1}{\sqrt{1-x^2}}\) (C) \(\frac{1}{\sqrt{1+x^2}}\) (D) \(\frac{x}{\sqrt{1+x^2}}\)
Let \(\tan^{-1} x = \theta\). Then \(\tan\theta = x\).
In a right triangle: opposite = \(x\), adjacent = 1, hypotenuse = \(\sqrt{1 + x^2}\).
So \(\sin\theta = \frac{x}{\sqrt{1 + x^2}}\).
Answer: (D)
In a right triangle: opposite = \(x\), adjacent = 1, hypotenuse = \(\sqrt{1 + x^2}\).
So \(\sin\theta = \frac{x}{\sqrt{1 + x^2}}\).
Answer: (D)
Q14. \(\sin^{-1}(1-x) - 2\sin^{-1} x = \frac{\pi}{2}\), then \(x\) is equal to
(A) 0, \(\frac{1}{2}\) (B) 1, \(\frac{1}{2}\) (C) 0 (D) \(\frac{1}{2}\)
(A) 0, \(\frac{1}{2}\) (B) 1, \(\frac{1}{2}\) (C) 0 (D) \(\frac{1}{2}\)
Rearranging: \(\sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1} x\).
Taking sine of both sides: \(1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right) = \cos(2\sin^{-1} x)\).
Let \(\sin^{-1} x = \alpha\), so \(\cos 2\alpha = 1 - 2\sin^2\alpha = 1 - 2x^2\).
Thus: \(1 - x = 1 - 2x^2\), giving \(2x^2 - x = 0\), i.e., \(x(2x - 1) = 0\).
So \(x = 0\) or \(x = \frac{1}{2}\).
Check \(x = \frac{1}{2}\): LHS = \(\sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \neq \frac{\pi}{2}\). Does not satisfy.
Check \(x = 0\): LHS = \(\sin^{-1} 1 - 0 = \frac{\pi}{2}\). Satisfies.
Answer: (C) \(x = 0\)
Taking sine of both sides: \(1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right) = \cos(2\sin^{-1} x)\).
Let \(\sin^{-1} x = \alpha\), so \(\cos 2\alpha = 1 - 2\sin^2\alpha = 1 - 2x^2\).
Thus: \(1 - x = 1 - 2x^2\), giving \(2x^2 - x = 0\), i.e., \(x(2x - 1) = 0\).
So \(x = 0\) or \(x = \frac{1}{2}\).
Check \(x = \frac{1}{2}\): LHS = \(\sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \neq \frac{\pi}{2}\). Does not satisfy.
Check \(x = 0\): LHS = \(\sin^{-1} 1 - 0 = \frac{\pi}{2}\). Satisfies.
Answer: (C) \(x = 0\)
Summary
Key Points — Chapter 2
| Function | Domain | Range (PVB) |
|---|---|---|
| \(\sin^{-1}\) | \([-1, 1]\) | \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) |
| \(\cos^{-1}\) | \([-1, 1]\) | \([0, \pi]\) |
| \(\tan^{-1}\) | \(\mathbf{R}\) | \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) |
| \(\cot^{-1}\) | \(\mathbf{R}\) | \((0, \pi)\) |
| \(\sec^{-1}\) | \(\mathbf{R} - (-1, 1)\) | \([0, \pi] - \{\frac{\pi}{2}\}\) |
| \(\csc^{-1}\) | \(\mathbf{R} - (-1, 1)\) | \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\) |
- \(\sin^{-1} x \neq (\sin x)^{-1}\). The notation \(\sin^{-1} x\) means the inverse function, NOT the reciprocal.
- For suitable domains: \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\), \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\), \(\csc^{-1} x + \sec^{-1} x = \frac{\pi}{2}\).
- The value of an inverse trigonometric function that lies in its principal value branch is called its principal value.
Historical Note
The study of trigonometry originated in ancient India. Aryabhata (476 A.D.), Brahmagupta (598 A.D.), Bhaskara I (600 A.D.), and Bhaskara II (1114 A.D.) all contributed significant results. The symbols \(\sin^{-1} x\), \(\cos^{-1} x\), etc., were suggested by astronomer Sir John F. W. Herschel (1813). The name of Thales (about 600 B.C.) is associated with height-and-distance problems solved using the principle of similar triangles.
Activity: Right Triangle Method for Simplification
Predict: If \(\theta = \tan^{-1}\frac{3}{4}\), can you find \(\sin\theta\) and \(\cos\theta\) without a calculator?
- Since \(\tan\theta = \frac{3}{4}\), draw a right triangle with opposite = 3 and adjacent = 4.
- By Pythagoras: hypotenuse = \(\sqrt{9 + 16} = 5\).
- Read off: \(\sin\theta = \frac{3}{5}\) and \(\cos\theta = \frac{4}{5}\).
- Now you can simplify: \(\sin(\tan^{-1}\frac{3}{4}) = \frac{3}{5}\) and \(\cos(\tan^{-1}\frac{3}{4}) = \frac{4}{5}\).
- Try this: what is \(\sin(2\tan^{-1}\frac{3}{4})\)?
Using \(\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}\).
Key Technique: The right triangle method works for all expressions of the form "trig function of inverse trig function". Assign sides based on the inner inverse function, compute the hypotenuse, then read off the outer function. This avoids complex algebraic identities and is especially useful for objective questions.
Common triples to remember: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25).
Competency-Based Questions
An engineer designs a ramp for wheelchair access. The ramp rises 1 m over a horizontal distance of \(d\) metres. The angle of inclination is \(\alpha = \tan^{-1}\left(\frac{1}{d}\right)\). Building codes require the angle to be no more than \(4.76^\circ\) (approximately \(\tan^{-1}\frac{1}{12}\)).
Q1. What minimum horizontal distance \(d\) is needed to satisfy the building code?
L3 ApplyThe angle must satisfy \(\tan^{-1}\frac{1}{d} \leq \tan^{-1}\frac{1}{12}\). Since \(\tan^{-1}\) is increasing, this gives \(\frac{1}{d} \leq \frac{1}{12}\), i.e., \(d \geq 12\) m. Minimum distance = 12 m.
Q2. Using the identity \(\sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}\), find the sine of the ramp angle when \(d = 12\).
L3 Apply\(\sin\alpha = \sin\left(\tan^{-1}\frac{1}{12}\right) = \frac{1/12}{\sqrt{1+(1/12)^2}} = \frac{1/12}{\sqrt{1+1/144}} = \frac{1/12}{\sqrt{145/144}} = \frac{1/12}{\frac{\sqrt{145}}{12}} = \frac{1}{\sqrt{145}} \approx 0.083\).
Q3. If two ramps of angles \(\alpha = \tan^{-1}\frac{1}{12}\) and \(\beta = \tan^{-1}\frac{1}{15}\) are joined, find the combined angle using the \(\tan^{-1}\) addition formula.
L4 Analyse\(\alpha + \beta = \tan^{-1}\left(\frac{\frac{1}{12}+\frac{1}{15}}{1-\frac{1}{12}\cdot\frac{1}{15}}\right) = \tan^{-1}\left(\frac{\frac{5+4}{60}}{1-\frac{1}{180}}\right) = \tan^{-1}\left(\frac{\frac{9}{60}}{\frac{179}{180}}\right) = \tan^{-1}\left(\frac{9 \times 180}{60 \times 179}\right) = \tan^{-1}\left(\frac{1620}{10740}\right) = \tan^{-1}\left(\frac{27}{179}\right) \approx 8.58^\circ\).
Q4. A student claims: "If \(\theta_1 + \theta_2 = \frac{\pi}{2}\), then \(\tan\theta_1 \cdot \tan\theta_2 = 1\)." Verify using inverse trig properties and decide if this is always true.
L5 EvaluateTrue. If \(\theta_1 + \theta_2 = \frac{\pi}{2}\), then \(\theta_2 = \frac{\pi}{2} - \theta_1\), so \(\tan\theta_2 = \tan(\frac{\pi}{2} - \theta_1) = \cot\theta_1 = \frac{1}{\tan\theta_1}\). Hence \(\tan\theta_1 \cdot \tan\theta_2 = 1\).
This is consistent with the property \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\), since \(\cot^{-1} x = \tan^{-1}\frac{1}{x}\) for \(x > 0\).
This is consistent with the property \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\), since \(\cot^{-1} x = \tan^{-1}\frac{1}{x}\) for \(x > 0\).
Assertion–Reason Questions
Assertion (A): \(\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{4}{5} = \frac{\pi}{2}\)
Reason (R): \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\) for \(x \in [-1, 1]\).
Reason (R): \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\) for \(x \in [-1, 1]\).
Answer: (a) — Since \(\sin^{-1}\frac{3}{5} = \cos^{-1}\frac{4}{5}\) (from the 3-4-5 right triangle), we get \(\cos^{-1}\frac{4}{5} + \sin^{-1}\frac{4}{5} = \frac{\pi}{2}\) by R. Both true, R explains A.
Assertion (A): \(\sin^{-1}(\sin 5) = 5\)
Reason (R): \(\sin^{-1}(\sin x) = x\) is valid only when \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Reason (R): \(\sin^{-1}(\sin x) = x\) is valid only when \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Answer: (d) — A is false: 5 radians is NOT in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \approx [-1.57, 1.57]\). We must first write \(\sin 5 = \sin(5 - 2\pi) \approx \sin(-1.283)\). Since \(-1.283 \in [-\frac{\pi}{2}, \frac{\pi}{2}]\), the answer is approximately \(-1.283\), not 5. R is true and explains why A fails.
Frequently Asked Questions
How to find principal values in NCERT exercises?
Identify the inverse trig function, recall its principal branch range, compute the matching angle, and ensure it lies within the principal branch. Verify with the forward trig function.
What exercises are in NCERT Class 12 Chapter 2?
Chapter 2 contains Exercise 2.1 (principal values), Exercise 2.2 (proving identities and simplifying), and a Miscellaneous Exercise with challenging combined problems.
How to solve 2tan-inverse problems?
Use identities: 2tan-inverse(x) = sin-inverse(2x/(1+x^2)) for |x| <= 1, or cos-inverse((1-x^2)/(1+x^2)) for x >= 0. Choose based on the target form.
What are common mistakes in inverse trig problems?
Forgetting principal value restrictions, using addition formula without checking xy < 1, confusing domains, and sign errors in negative argument identities.
How to approach miscellaneous exercises on inverse trig?
Identify needed properties, convert to a common form, work from one side for proofs, and isolate the inverse trig term for equation problems.
Frequently Asked Questions — Inverse Trigonometric Functions
What is Exercises and Summary - Inverse Trigonometric Functions in NCERT Class 12 Mathematics?
Exercises and Summary - Inverse Trigonometric Functions is a key concept covered in NCERT Class 12 Mathematics, Chapter 2: Inverse Trigonometric Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Exercises and Summary - Inverse Trigonometric Functions step by step?
To solve problems on Exercises and Summary - Inverse Trigonometric Functions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Inverse Trigonometric Functions?
The essential formulas of Chapter 2 (Inverse Trigonometric Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Exercises and Summary - Inverse Trigonometric Functions important for the Class 12 board exam?
Exercises and Summary - Inverse Trigonometric Functions is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Exercises and Summary - Inverse Trigonometric Functions?
Common mistakes in Exercises and Summary - Inverse Trigonometric Functions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Exercises and Summary - Inverse Trigonometric Functions?
End-of-chapter NCERT exercises for Exercises and Summary - Inverse Trigonometric Functions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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