Chapter 3: Trigonometry
1. The value of \(\tan^2 \theta \sec^2 \theta (\cot^2 \theta - \cos^2 \theta)\) is
Solution \((\cot^2\theta - \cos^2\theta) = \frac{\cos^2\theta}{\sin^2\theta} - \cos^2\theta = \cos^2\theta(\csc^2\theta-1) = \cos^2\theta \cot^2\theta\).
Expr = \(\tan^2\theta \sec^2\theta \cos^2\theta \cot^2\theta = (\tan^2\theta \cot^2\theta)(\sec^2\theta \cos^2\theta) = 1 \cdot 1 = 1\).
Expr = \(\tan^2\theta \sec^2\theta \cos^2\theta \cot^2\theta = (\tan^2\theta \cot^2\theta)(\sec^2\theta \cos^2\theta) = 1 \cdot 1 = 1\).
2. Value of \(\cot 5^\circ \cot 10^\circ \dots \cot 85^\circ\) is
Solution Terms can be paired: \(\cot 5^\circ \cot 85^\circ = \cot 5^\circ \tan 5^\circ = 1\).
The middle term is \(\cot 45^\circ = 1\). The product of all pairs is 1.
The middle term is \(\cot 45^\circ = 1\). The product of all pairs is 1.
3. Value of \(\sin 10^\circ + \sin 20^\circ + \sin 30^\circ + \dots + \sin 360^\circ\) is
Solution \(\sin(180^\circ + \theta) = -\sin \theta\).
Terms cancel out in pairs: \(\sin 10^\circ + \sin 190^\circ = 0\), etc.
\(\sin 180^\circ = 0, \sin 360^\circ = 0\). Total sum is 0.
Terms cancel out in pairs: \(\sin 10^\circ + \sin 190^\circ = 0\), etc.
\(\sin 180^\circ = 0, \sin 360^\circ = 0\). Total sum is 0.
4. If \(\tan A = \frac{1}{2}\) and \(\tan B = \frac{1}{3}\), then value of \(A + B\) is
Solution \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1/2 + 1/3}{1 - 1/6} = \frac{5/6}{5/6} = 1\).
\(\tan(A+B) = 1 \Rightarrow A+B = \frac{\pi}{4}\).
\(\tan(A+B) = 1 \Rightarrow A+B = \frac{\pi}{4}\).
5. If \(\sin 2\theta + \sin 2\phi = 1/2, \cos 2\theta + \cos 2\phi = 3/2\), then value of \(\cos^2(\theta - \phi)\) is
Solution Squaring and adding:
\(2 + 2\cos(2\theta - 2\phi) = 1/4 + 9/4 = 10/4 = 5/2\).
\(2(1 + \cos 2(\theta-\phi)) = 5/2\). Using \(1+\cos 2x = 2\cos^2 x\):
\(2(2\cos^2(\theta-\phi)) = 5/2 \Rightarrow 4\cos^2(\theta-\phi) = 5/2 \Rightarrow \cos^2(\theta-\phi) = 5/8\).
\(2 + 2\cos(2\theta - 2\phi) = 1/4 + 9/4 = 10/4 = 5/2\).
\(2(1 + \cos 2(\theta-\phi)) = 5/2\). Using \(1+\cos 2x = 2\cos^2 x\):
\(2(2\cos^2(\theta-\phi)) = 5/2 \Rightarrow 4\cos^2(\theta-\phi) = 5/2 \Rightarrow \cos^2(\theta-\phi) = 5/8\).
6. If \(0 < \theta < 360^\circ\), then solutions of \(\cos \theta = -1/2\) are
Solution \(\cos \theta\) is negative in 2nd and 3rd quadrants. Ref angle is \(60^\circ\).
Q2: \(180-60=120^\circ\). Q3: \(180+60=240^\circ\).
Q2: \(180-60=120^\circ\). Q3: \(180+60=240^\circ\).
7. If \(\tan \theta = -\frac{1}{\sqrt{3}}\), then general solution of the equation is
Solution \(\tan \theta = \tan(-\frac{\pi}{6})\). General solution: \(\theta = n\pi + \alpha\).
Here \(\alpha = -\pi/6\), so \(\theta = n\pi - \frac{\pi}{6}\).
Here \(\alpha = -\pi/6\), so \(\theta = n\pi - \frac{\pi}{6}\).
8. If \(2 \tan^2 \theta = \sec^2 \theta\), then general value of \(\theta\) are
Solution \(2\tan^2\theta = 1 + \tan^2\theta \Rightarrow \tan^2\theta = 1 \Rightarrow \tan \theta = \pm 1\).
\(\theta = n\pi \pm \frac{\pi}{4}\).
\(\theta = n\pi \pm \frac{\pi}{4}\).
9. If \(\sin 5x + \sin 3x + \sin x = 0\) and \(0 \le x \le \pi/2\), then value of x is
Solution \((\sin 5x + \sin x) + \sin 3x = 2\sin 3x \cos 2x + \sin 3x = \sin 3x (2\cos 2x + 1) = 0\).
\(\sin 3x=0 \Rightarrow x=0, \pi/3\). \(\cos 2x = -1/2 \Rightarrow 2x = 2\pi/3 \Rightarrow x=\pi/3\).
Among options, \(\pi/3\) is the correct non-zero solution.
\(\sin 3x=0 \Rightarrow x=0, \pi/3\). \(\cos 2x = -1/2 \Rightarrow 2x = 2\pi/3 \Rightarrow x=\pi/3\).
Among options, \(\pi/3\) is the correct non-zero solution.
10. If \(y = \frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}\), then value of \(\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}\) is
Solution Multiplying num/den of given 'y' by appropriate conjugate or solving the target expression by rationalization shows it is equal to \(y\).
Example step: \(\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha} \cdot \frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}\) results in y.
Example step: \(\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha} \cdot \frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}\) results in y.
11. The number of solution of \(\tan x + \sec x = 2\cos x\) in \((0, 2\pi)\) is
Solution \(\frac{\sin x + 1}{\cos x} = 2\cos x \Rightarrow \sin x + 1 = 2(1-\sin^2 x)\).
\(2\sin^2 x + \sin x - 1 = 0 \Rightarrow (2\sin x - 1)(\sin x + 1) = 0\).
\(\sin x = 1/2\) (2 solutions in range) or \(\sin x = -1\) (1 solution).
At \(\sin x = -1, \cos x = 0\), eq is undefined. So only 2 valid solutions.
\(2\sin^2 x + \sin x - 1 = 0 \Rightarrow (2\sin x - 1)(\sin x + 1) = 0\).
\(\sin x = 1/2\) (2 solutions in range) or \(\sin x = -1\) (1 solution).
At \(\sin x = -1, \cos x = 0\), eq is undefined. So only 2 valid solutions.
12. If \(\sin A = \frac{3}{5}, 0 < A < \frac{\pi}{2}\) and \(\cos B = \frac{-12}{13}, \pi < B < \frac{3\pi}{2}\), then value of \(\sin(A-B)\) is
Solution \(\cos A = 4/5, \sin B = -5/13\).
\(\sin(A-B) = \sin A \cos B - \cos A \sin B\)
\(= (3/5)(-12/13) - (4/5)(-5/13) = -36/65 + 20/65 = -16/65\).
\(\sin(A-B) = \sin A \cos B - \cos A \sin B\)
\(= (3/5)(-12/13) - (4/5)(-5/13) = -36/65 + 20/65 = -16/65\).
13. Value of \(\tan 15^\circ \cdot \tan 45^\circ \tan 75^\circ\) is
Solution \(\tan 75^\circ = \cot 15^\circ\).
Expr \(= \tan 15^\circ \cdot 1 \cdot \cot 15^\circ = 1\).
Expr \(= \tan 15^\circ \cdot 1 \cdot \cot 15^\circ = 1\).
14. Value of \(\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)\) is
Solution \(\cos\frac{7\pi}{8} = -\cos\frac{\pi}{8}\), etc.
Expr \(= (1-\cos^2\frac{\pi}{8})(1-\cos^2\frac{3\pi}{8}) = \sin^2\frac{\pi}{8} \sin^2\frac{3\pi}{8}\)
\(= \sin^2\frac{\pi}{8} \cos^2\frac{\pi}{8} = \frac{1}{4}(2\sin\frac{\pi}{8}\cos\frac{\pi}{8})^2 = \frac{1}{4}(\sin\frac{\pi}{4})^2 = \frac{1}{4}(\frac{1}{2}) = \frac{1}{8}\).
Expr \(= (1-\cos^2\frac{\pi}{8})(1-\cos^2\frac{3\pi}{8}) = \sin^2\frac{\pi}{8} \sin^2\frac{3\pi}{8}\)
\(= \sin^2\frac{\pi}{8} \cos^2\frac{\pi}{8} = \frac{1}{4}(2\sin\frac{\pi}{8}\cos\frac{\pi}{8})^2 = \frac{1}{4}(\sin\frac{\pi}{4})^2 = \frac{1}{4}(\frac{1}{2}) = \frac{1}{8}\).
15. The large hand of a clock is 42 cm long. How much distance does its extremity move in 20 minutes?
Solution Angle in 20 min = 120 degrees = \(2\pi/3\) radians.
Arc length \(l = r\theta = 42 \times \frac{2}{3} \times \frac{22}{7} = 42 \times \frac{44}{21} = 2 \times 44 = 88\).
Arc length \(l = r\theta = 42 \times \frac{2}{3} \times \frac{22}{7} = 42 \times \frac{44}{21} = 2 \times 44 = 88\).
16. The angle in radian through which a pendulum swings and its length is 75 cm and tip describes an arc of length 21 cm, is
Solution \(l = 21, r = 75\).
\(\theta = l/r = 21/75 = 7/25\) radians.
\(\theta = l/r = 21/75 = 7/25\) radians.
17. The length of an arc of a circle of radius 3 cm, if the angle subtended at the centre is \(30^\circ\) is (\(\pi = 3.14\))
Solution \(\theta = 30^\circ = \pi/6\).
\(l = r\theta = 3 \times (\pi/6) = \pi/2 = 3.14/2 = 1.57\) cm.
\(l = r\theta = 3 \times (\pi/6) = \pi/2 = 3.14/2 = 1.57\) cm.
18. A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm. The angle subtended by the arc at the centre is
Solution Length of wire = \(2\pi(7) = 14\pi\).
New radius \(r=12\). \(l = 14\pi\).
\(\theta = l/r = 14\pi/12 = 7\pi/6\).
Degrees: \((7 \times 180)/6 = 210^\circ\).
New radius \(r=12\). \(l = 14\pi\).
\(\theta = l/r = 14\pi/12 = 7\pi/6\).
Degrees: \((7 \times 180)/6 = 210^\circ\).
19. A circular wire of radius 3 cm is cut and bent so as to lie along the circumference of a hoop whose radius is 48 cm. The angle in degrees which is subtended at the centre of hoop is
Solution Length \(l = 2\pi(3) = 6\pi\). Hoop radius \(R = 48\).
\(\theta = 6\pi/48 = \pi/8\) radians.
Degrees: \(180/8 = 22.5^\circ\).
\(\theta = 6\pi/48 = \pi/8\) radians.
Degrees: \(180/8 = 22.5^\circ\).
20. The radius of the circle in which a central angle of \(60^\circ\) intercepts an arc of length 37.4 cm is (Use \(\pi = 22/7\))
Solution \(\theta = 60^\circ = \pi/3\). \(l = 37.4\).
\(r = l/\theta = 37.4 / (\pi/3) = 37.4 \times 3 \times 7 / 22\).
\(37.4 \times 21 / 22 = 1.7 \times 21 = 35.7\) cm.
\(r = l/\theta = 37.4 / (\pi/3) = 37.4 \times 3 \times 7 / 22\).
\(37.4 \times 21 / 22 = 1.7 \times 21 = 35.7\) cm.