Every Book Back multiple-choice question from Trigonometry (10th Standard Mathematics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
The value of \(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\) is equal to
- A. \(\tan^2 \theta\)
- B. 1Correct
- C. \(\cot^2 \theta\)
- D. 0
Explanation. Since \(1 + \tan^2 \theta = \sec^2 \theta\), the expression becomes \(\sin^2 \theta + \frac{1}{\sec^2 \theta} = \sin^2 \theta + \cos^2 \theta\). By the fundamental trigonometric identity, this sum is equal to 1.
Q2
\(\tan \theta \csc^2 \theta - \tan \theta\) is equal to
- A. \(\sec \theta\)
- B. \(\cot^2 \theta\)
- C. \(\sin \theta\)
- D. \(\cot \theta\)Correct
Explanation. Factoring out \(\tan \theta\) gives \(\tan \theta (\csc^2 \theta - 1)\). Using the identity \csc^2 \theta - 1 = \cot^2 \theta\(, we get \)\tan \theta \cot^2 \theta = \tan \theta \cdot \frac{1}{\tan^2 \theta} = \cot \theta.
Q3
If \((\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2 \alpha + \cot^2 \alpha\), then the value of \(k\) is equal to
- A. 9
- B. 7Correct
- C. 5
- D. 3
Explanation. Expanding the squares yields \sin^2 \alpha + \csc^2 \alpha + 2 + \cos^2 \alpha + \sec^2 \alpha + 2\(. Using \)\sin^2 \alpha + \cos^2 \alpha = 1\(, \)\csc^2 \alpha = 1 + \cot^2 \alpha\(, and \sec^2 \alpha = 1 + \tan^2 \alpha\), we simplify to \(7 + \tan^2 \alpha + \cot^2 \alpha\), so \(k = 7\).
Q4
If \(\sin \theta + \cos \theta = a\) and \(\sec \theta + \csc \theta = b\), then the value of \(b(a^2 - 1)\) is equal to
- A. \(2a\)Correct
- B. \(3a\)
- C. 0
- D. \(2ab\)
Explanation. Squaring the first equation gives \(a^2 = 1 + 2\sin \theta \cos \theta\), so \(a^2 - 1 = 2\sin \theta \cos \theta\). Expressing \(b\) as \(\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \frac{a}{\sin \theta \cos \theta}\), we find \(b(a^2 - 1) = 2a\).
Q5
If \(5x = \sec \theta\) and \(\frac{5}{y} = \tan \theta\), then \(x^2 - \frac{1}{y^2}\) is equal to
- A. 25
- B. \(\frac{1}{25}\)Correct
- C. 5
- D. 1
Explanation. We have \(x = \frac{\sec \theta}{5}\) and \(\frac{1}{y} = \frac{\tan \theta}{5}\). Squaring both and subtracting gives \(x^2 - \frac{1}{y^2} = \frac{\sec^2 \theta - \tan^2 \theta}{25}\). Since \(\sec^2 \theta - \tan^2 \theta = 1\), this equals \(\frac{1}{25}\).
Q6
If \(\sin \theta = \cos \theta\), then \(2\tan^2 \theta + \sin^2 \theta - 1\) is equal to
- A. \(-\frac{3}{2}\)
- B. \(\frac{3}{2}\)Correct
- C. \(\frac{2}{3}\)
- D. \(-\frac{2}{3}\)
Explanation. Since \(\sin \theta = \cos \theta\), we have \(\theta = 45^\circ\) for acute \(\theta\). Substituting this value, the expression becomes \(2\tan^2 45^\circ + \sin^2 45^\circ - 1 = 2(1)^2 + (\frac{1}{\sqrt{2}})^2 - 1 = 2 + \frac{1}{2} - 1 = \frac{3}{2}\).
Q7
If \(x = a\tan \theta\) and \(y = b\sec \theta\), then
- A. \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\)Correct
- B. \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
- C. \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
- D. \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0\)
Explanation. From the given equations, we have \(\tan \theta = \frac{x}{a}\) and \(\sec \theta = \frac{y}{b}\). Substituting these into the fundamental trigonometric identity \(\sec^2 \theta - \tan^2 \theta = 1\) yields \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\).
Q8
\((1 + \tan \theta + \sec \theta)(1 + \cot \theta - \csc \theta)\) is equal to
- A. 0
- B. 1
- C. 2Correct
- D. -1
Explanation. Converting all terms to sine and cosine gives \(\left(\frac{\sin \theta + \cos \theta + 1}{\cos \theta}\right)\left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) = \frac{(\sin \theta + \cos \theta)^2 - 1}{\sin \theta \cos \theta} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = 2\).
Q9
\(a\cot \theta + b\csc \theta = p\) and \(b\cot \theta + a\csc \theta = q\), then \(p^2 - q^2\) is equal to
- A. \(a^2 - b^2\)
- B. \(b^2 - a^2\)Correct
- C. \(a^2 + b^2\)
- D. \(b - a\)
Explanation. Calculating \(p^2 - q^2\) yields \((a^2 \cot^2 \theta + b^2 \csc^2 \theta + 2ab\cot\theta\csc\theta) - (b^2 \cot^2 \theta + a^2 \csc^2 \theta + 2ab\cot\theta\csc\theta) = (b^2 - a^2)(\csc^2 \theta - \cot^2 \theta)\). Since \(\csc^2 \theta - \cot^2 \theta = 1\), we get \(b^2 - a^2\).
Q10
If the ratio of the height of a tower and the length of its shadow is \(\sqrt{3} : 1\), then the angle of elevation of the sun has measure
- A. \(45^\circ\)
- B. \(30^\circ\)
- C. \(90^\circ\)
- D. \(60^\circ\)Correct
Explanation. Let \(h\) be the height of the tower and \(s\) be the length of its shadow. We are given \(\frac{h}{s} = \frac{\sqrt{3}}{1}\). Since \(\tan \theta = \frac{h}{s}\), we have \(\tan \theta = \sqrt{3}\), which means \(\theta = 60^\circ\).
Q11
An electric pole subtends an angle of \(30^\circ\) at a point on the same level as its foot. At a second point \(b\) metres above the first, the depression of the foot of the pole is \(60^\circ\). The height of the pole (in metres) is equal to
- A. \(b\sqrt{3}\)
- B. \(\frac{b}{3}\)Correct
- C. \(\frac{b}{2}\)
- D. \(\frac{b}{\sqrt{3}}\)
Explanation. Let \(h\) be the height of the pole and \(d\) be the distance to its foot. From the first point, \(d = h\sqrt{3}\). From the second point, \(d = \frac{b}{\sqrt{3}}\). Equating these yields \(h\sqrt{3} = \frac{b}{\sqrt{3}}\), which simplifies to \(h = \frac{b}{3}\).
Q12
A tower is \(60\text{ m}\) high. Its shadow reduces by \(x\text{ metres}\) when the angle of elevation of the sun increases from \(30^\circ\) to \(45^\circ\), then \(x\) is equal to
- A. \(41.92\text{ m}\)
- B. \(43.92\text{ m}\)Correct
- C. \(43\text{ m}\)
- D. \(45.6\text{ m}\)
Explanation. Let the shadow lengths at \(45^\circ\) and \(30^\circ\) be \(y\) and \(y + x\) respectively. We have \(y = 60\cot 45^\circ = 60\text{ m}\), and \(y + x = 60\cot 30^\circ = 60\sqrt{3}\text{ m}\). Thus, \(x = 60(\sqrt{3} - 1) \approx 60(0.732) = 43.92\text{ m}\).
Q13
The angles of depression of the top and bottom of a \(20\text{ m}\) tall building from the top of a multistoried building are \(30^\circ\) and \(60^\circ\) respectively. The height of the multistoried building and the distance between the two buildings (in metres) are
- A. \(20,\ 10\sqrt{3}\)
- B. \(30,\ 5\sqrt{3}\)
- C. \(20,\ 10\)
- D. \(30,\ 10\sqrt{3}\)Correct
Explanation. Let \(H\) be the multistoried building's height and \(D\) be the distance. Then, \(D = \frac{H}{\sqrt{3}}\) and \(D = (H - 20)\sqrt{3}\). Equating these gives \(H = 3(H - 20) \implies H = 30\text{ m}\), and \(D = \frac{30}{\sqrt{3}} = 10\sqrt{3}\text{ m}\).
Q14
Two persons are standing \(x\text{ metres}\) apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is
- A. \(2x\)
- B. \(\frac{x}{2\sqrt{2}}\)Correct
- C. \(\frac{x}{2}\)
- D. \(2\sqrt{2}x\)
Explanation. Let the heights be \(2h\) and \(h\). From the midpoint \(\frac{x}{2}\), the tangents of the complementary angles are \(\frac{4h}{x}\) and \(\frac{2h}{x}\). Multiplying them gives \(\frac{8h^2}{x^2} = \tan \theta \cot \theta = 1 \implies h^2 = \frac{x^2}{8} \implies h = \frac{x}{2\sqrt{2}}\).
Q15
The angle of elevation of a cloud from a point \(h\) metres above a lake is \(\beta\). The angle of depression of its reflection in the lake is \(45^\circ\). The height of location of the cloud from the lake is
- A. \(\frac{h(1 + \tan \beta)}{1 - \tan \beta}\)Correct
- B. \(\frac{h(1 - \tan \beta)}{1 + \tan \beta}\)
- C. \(h\tan(45^\circ - \beta)\)
- D. none of these
Explanation. Let \(H\) be the cloud's height. The horizontal distance to the cloud is \(d = (H - h)\cot \beta\) and \(d = H + h\) because the reflection angle is \(45^\circ\). Thus, \(H + h = \frac{H - h}{\tan \beta} \implies H = \frac{h(1 + \tan \beta)}{1 - \tan \beta}\).
