CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 02

Multiple Choice Questions

Question. If \(\tan A = \frac{4}{3}\), the value of \(\sin A\) is
(a) \(\frac{4}{5}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{5}{3}\)
(d) \(\frac{7}{5}\)
Answer: (a)
Explanation: \(\tan A = \frac{4}{3}\)
\(\Rightarrow \tan^2 A = \frac{16}{9} \Rightarrow \cot^2 A = \frac{9}{16}\)
\(\therefore \csc^2 A - 1 = \frac{9}{16} \Rightarrow \csc^2 A = \frac{25}{16}\)
\(\Rightarrow \csc A = \frac{5}{4} \Rightarrow \sin A = \frac{4}{5}\)

Question. If \(3 \tan A = 4\), then the value of \(\frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A}\) is:
(a) 4
(b) \(\frac{11}{15}\)
(c) \(\frac{7}{15}\)
(d) 3
Answer: (d)
Explanation: \(3 \tan A = 4 \Rightarrow \tan A = \frac{4}{3}\)
\(\therefore \frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A}\)
Divide numerator and denominator by \(\cos A\)
\(\Rightarrow \frac{\frac{3 \sin A}{\cos A} + \frac{2 \cos A}{\cos A}}{\frac{3 \sin A}{\cos A} - \frac{2 \cos A}{\cos A}} = \frac{3 \tan A + 2}{3 \tan A - 2}\)
\(\Rightarrow \frac{3 \times \frac{4}{3} + 2}{3 \times \frac{4}{3} - 2} = \frac{4 + 2}{4 - 2} = \frac{6}{2} = 3\)

Question. If \(\sin \theta + \cos \theta = \sqrt{2} \cos \theta\), (\(\theta \neq 90^\circ\)) then the value of \(\tan \theta\) is:
(a) \(\sqrt{2} - 1\)
(b) \(\sqrt{2} + 1\)
(c) \(\sqrt{2}\)
(d) \(-\sqrt{2}\)
Answer: (a) 

Question. Given that \(\sin \alpha = \frac{\sqrt{3}}{2}\) and \(\cos \beta = 0\), then the value of \(\beta - \alpha\) is:
(a) \(0^\circ\)
(b) \(90^\circ\)
(c) \(60^\circ\)
(d) \(30^\circ\)
Answer: (d) 

Question. If \(\sin (A + B) = \cos (A - B) = 1\), then
(a) \(A = B = 0\)
(b) \(A = B = 45^\circ\)
(c) \(A = 60^\circ, B = 30^\circ\)
(d) None of these
Answer: (b)
Explanation: \(\sin (A + B) = \cos (A - B) = 1\)
\(\Rightarrow \sin (A + B) = 1 \Rightarrow \sin (A + B) = \sin 90^\circ\)
\(\Rightarrow A + B = 90^\circ\) ...(i)
\(\Rightarrow \cos (A - B) = 1 \Rightarrow \cos (A - B) = \cos 0^\circ\)
\(\Rightarrow A - B = 0\) or \(A = B\)
Putting in (i), \(2A = 90^\circ \Rightarrow A = 45^\circ\)
\(A = B = 45^\circ\)

Question. If \(\cos A = \frac{5}{13}\), find the value of \(\tan A + \cot A\)
(a) \(\frac{169}{60}\)
(b) \(\frac{12}{13}\)
(c) 1
(d) \(\frac{60}{169}\)
Answer: (a)
Explanation: \(\cos A = \frac{5}{13}\)
\(\Rightarrow \cos^2 A = \frac{25}{169}, 1 - \sin^2 A = \frac{25}{169}\)
\(\Rightarrow \sin^2 A = \frac{144}{169}\) or \(\sin A = \frac{12}{13}\)
\(\Rightarrow \tan A = \frac{\sin A}{\cos A} = \frac{12/13}{5/13} = \frac{12}{5}\)
\(\Rightarrow \cot A = \frac{1}{\tan A} = \frac{5}{12}\)
\(\tan A + \cot A = \frac{12}{5} + \frac{5}{12} = \frac{169}{60}\)

Question. If \(5x = \sec \theta\) and \(\frac{5}{x} = \tan \theta\) then find the value of \(5(x^2 - \frac{1}{x^2})\)
(a) 5
(b) \(\frac{1}{5}\)
(c) \(\frac{2}{5}\)
(d) 0
Answer: (b)
Explanation: \(5x = \sec \theta\), \(\frac{5}{x} = \tan \theta\)
Squaring, \(25x^2 = \sec^2 \theta\), \(\frac{25}{x^2} = \tan^2 \theta\)
Subtracting,
\(\Rightarrow 25x^2 - \frac{25}{x^2} = \sec^2 \theta - \tan^2 \theta\)
\(\Rightarrow 25(x^2 - \frac{1}{x^2}) = 1\)
\(\Rightarrow 5(x^2 - \frac{1}{x^2}) = \frac{1}{5}\)

Question. If \(\sin A + \sin^2 A = 1\), then the value of the expression \((\cos^2 A + \cos^4 A)\) is:
(a) 1
(b) \(\frac{1}{2}\)
(c) 2
(d) 3 
Answer: (a)
Explanation: We know that \(\sin^2 \theta + \cos^2 \theta = 1\) ....(i)
Given: \(\sin A + \sin^2 A = 1\)
\(\Rightarrow \sin A = 1 - \sin^2 A\)
\(\Rightarrow \sin A = \cos^2 A\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
Squaring both sides
\(\Rightarrow \sin^2 A = \cos^4 A\)
\(\Rightarrow 1 - \cos^2 A = \cos^4 A\) [using (i)]
\(\Rightarrow \cos^2 A + \cos^4 A = 1\)

Question. The value of \((1 + \cos \theta) (1 - \cos \theta) \csc^2 \theta =\)
(a) 0
(b) 1
(c) \(\cos^2 \theta\)
(d) \(\sin^2 \theta\)
Answer: (b)
Explanation: \((1 + \cos \theta) (1 - \cos \theta) \csc^2 \theta\)
\(= (1 - \cos^2 \theta) \csc^2 \theta\)
\(= \sin^2 \theta \csc^2 \theta = 1\)

Question. If \(\sin \theta - \cos \theta = 0\), then the value of \((\sin^4 \theta + \cos^4 \theta)\) is:
(a) 1
(b) \(\frac{3}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{4}\) 
Answer: (c)
Explanation: We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
Given: \((\sin \theta - \cos \theta) = 0\)
\(\Rightarrow \sin \theta = \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = 1\)
\(\Rightarrow \tan \theta = 1 \Rightarrow \tan \theta = \tan 45^\circ\)
\(\Rightarrow \theta = 45^\circ\)
Now, \(\sin^4 \theta + \cos^4 \theta = \sin^4 45^\circ + \cos^4 45^\circ\)
\(= \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)

Question. If \(\theta = 45^\circ\) then \(\sec \theta \cot \theta - \csc \theta \tan \theta\) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a)
Explanation: \(\sec \theta \cot \theta - \csc \theta \tan \theta\)
\(= \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta}\)
\(= \frac{1}{\sin \theta} - \frac{1}{\cos \theta}\)
{At \(\theta = 45^\circ\), \(\sin \theta = \cos \theta\)}
\(= \frac{1}{\sin \theta} - \frac{1}{\sin \theta} = 0\)

Question. If \(x = a \sin \theta\) and \(y = a \cos \theta\), then the value of \(x^2 + y^2\) is
(a) \(a\)
(b) \(a^2\)
(c) 1
(d) \(b^2\)
Answer: (b)
Explanation: \(x = a \sin \theta, y = a \cos \theta\)
Squaring, \(x^2 = a^2 \sin^2 \theta, y^2 = a^2 \cos^2 \theta\)
Adding \(x^2 + y^2 = a^2 \sin^2 \theta + a^2 \cos^2 \theta\)
\(= a^2 (\sin^2 \theta + \cos^2 \theta) = a^2\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]

Question. \(4 \tan^2 A - 4 \sec^2 A\) is equal to:
(a) 2
(b) 3
(c) 4
(d) -4
Answer: (d)
Explanation: \(4 \tan^2 A - 4 \sec^2 A\)
\(= -4 (\sec^2 A - \tan^2 A)\)
\(= -4 \times 1 = -4\)

Question. If \(3 \cos \theta = 1\), then \(\csc \theta\) is equal to:
(a) \(2\sqrt{2}\)
(b) \(\frac{3}{2\sqrt{2}}\)
(c) \(\frac{2\sqrt{3}}{3}\)
(d) \(\frac{4}{3\sqrt{2}}\)
Answer: (b)
Explanation: \(3 \cos \theta = 1\) gives \(\cos \theta = \frac{1}{3}\)
Here \(B = 1, H = 3\)
\(\therefore P = \sqrt{H^2 - B^2} = \sqrt{9 - 1} = \sqrt{8} = 2\sqrt{2}\)
\(\csc \theta = \frac{H}{P} = \frac{3}{2\sqrt{2}}\)

Question. If \(\csc \theta - \cot \theta = \frac{1}{3}\), then the value of \(\csc \theta + \cot \theta\) is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c)
Explanation: We know that: \(\csc^2 \theta - \cot^2 \theta = 1\)
i.e. \((\csc \theta + \cot \theta)(\csc \theta - \cot \theta) = 1\)
\(\Rightarrow \csc \theta + \cot \theta = \frac{1}{\csc \theta - \cot \theta} = \frac{1}{1/3} = 3\)

Question. If \(2 \sin 2\theta = \sqrt{3}\), then the value of \(\theta\) is:
(a) \(90^\circ\)
(b) \(30^\circ\)
(c) \(60^\circ\)
(d) \(45^\circ\)
Answer: (b)
Explanation: \(2 \sin 2\theta = \sqrt{3} \Rightarrow \sin 2\theta = \frac{\sqrt{3}}{2}\)
\(\Rightarrow \sin 2\theta = \sin 60^\circ\)
\(\Rightarrow 2\theta = 60^\circ \Rightarrow \theta = 30^\circ\)

Question. If the height and length of the shadow of a man are the same, then the angle of elevation of the sun is:
(a) \(45^\circ\)
(b) \(60^\circ\)
(c) \(90^\circ\)
(d) \(120^\circ\) 
Answer: (a)
Explanation: Let \(AB\) be the height of a man and \(BC\) be the shadow of a man.
\(AB = BC\) [Given]
In right angled \(\Delta ABC\), \(\tan \theta = \frac{AB}{BC}\)
\(\Rightarrow \frac{AB}{AB} = \tan \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = 45^\circ\).

Write True False

Question. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection. 
Answer: False.
Explanation: Let a man be standing on a platform at point P, 3 m above the surface of the lake a cloud is observed at point C. Let the height of the cloud from the surface of the platform be \(h\). In \(\Delta MPC\), \(\tan \theta_1 = \frac{CM}{PM} = \frac{h}{PM}\). In \(\Delta LPM\), \(\tan \theta_2 = \frac{LM}{PM} = \frac{OL + OM}{PM} = \frac{h + 6}{PM}\). Hence \(\theta_1 \neq \theta_2\).

Question. \(\cos \theta = \frac{a^2 + b^2}{2ab}\), where \(a\) and \(b\) are two distinct numbers such that \(ab > 0\). 
Answer: False.
Explanation: We know that \((a - b)^2 > 0\) [As square of any number is positive]
\(\Rightarrow a^2 + b^2 - 2ab > 0 \Rightarrow a^2 + b^2 > 2ab \Rightarrow \frac{a^2 + b^2}{2ab} > 1\)
But \(\cos \theta = \frac{a^2 + b^2}{2ab}\) \(\Rightarrow \cos \theta > 1\)
which is not possible, since, \(-1 \leq \cos \theta \leq 1\). Hence, \(\cos \theta \neq \frac{a^2 + b^2}{2ab}\).

Question. The angle of elevation of the top of a tower is \(30^\circ\). If the height of the tower is doubled, then the angle of elevation of its top will also be doubled. 
Answer: False.
Explanation: \(\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\). Let \(AC\) be the tower with height '\(h\)' and \(BC = x\). In \(\Delta ABC\), \(\tan 30^\circ = \frac{AC}{BC} = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x}\) ....(i). When height of tower is doubled i.e., \(A'C' = 2h\). In \(\Delta A'B'C'\), \(\tan \theta = \frac{A'C'}{B'C'} = \frac{2h}{x} = 2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}\). But \(\tan 60^\circ = \sqrt{3} \neq \frac{2}{\sqrt{3}}\). So \(\theta \neq 60^\circ\). Hence, the required angle is not doubled.

Question. If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains unchanged.
Answer: True.
Explanation: Let \(AC\) be the tower of height \(h\) and the distance of the point of observation from its foot be \(x\). In \(\Delta ABC\), \(\tan \theta_1 = \frac{AC}{BC} = \frac{h}{x}\) ....(i). Now, if the height of the tower is increased by 10%, new height \(h' = h + 10\% \text{ of } h = h + \frac{h}{10} = \frac{11h}{10}\). Distance of point of observation from its foot is also increased by 10%. New distance \(x' = x + \frac{10x}{100} = x + \frac{x}{10} = \frac{11x}{10}\). In \(\Delta PQR\), \(\tan \theta_2 = \frac{PR}{QR} = \frac{h'}{x'} = \frac{11h/10}{11x/10} = \frac{h}{x}\) ....(ii). From eqn (i) and (ii), \(\tan \theta_1 = \tan \theta_2 \Rightarrow \theta_1 = \theta_2\). Hence, the required angle of elevation of its top remains unchanged.

Fill in the Blanks

Fill in the blanks/tables with suitable information:

Question. Simplest form of \(\frac{1 + \tan^2 A}{1 + \cot^2 A}\) is .................... . 
Answer: \(\tan^2 A\)
Explanation: \(\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A\).

Question. If \(\tan A = 1\), then \(2 \sin A \cos A = ....................\) 
Answer: 1
Explanation: Since, \(\tan A = 1\), \(\therefore A = 45^\circ\). \(\therefore 2 \sin A \cos A = 2 \times \sin 45^\circ \times \cos 45^\circ = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1\).

Question. In the figure, the angles of depressions from the observing positions \(O_1\) and \(O_2\), respectively of the object A are ...................., .................... . 
Answer: \(30^\circ, 45^\circ\)
Explanation: Angle of depression from \(O_1\) is \(\angle O_2O_1A = 90^\circ - \angle AO_1C = 90^\circ - 60^\circ = 30^\circ\). Angle of depressions from \(O_2\) is \(\angle XO_2A = 45^\circ\).

Question. The value of \((\sin^2 \theta + \frac{1}{1 + \tan^2 \theta})\) is .................... 
Answer: 1
Explanation: \(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta} = \sin^2 \theta + \frac{1}{\sec^2 \theta} = \sin^2 \theta + \cos^2 \theta = 1\).

Question. Simplest form of \((1 - \cos^2 A)(1 + \cot^2 A)\) is .................... .
Answer: 1
Explanation: \((1 - \cos^2 A)(1 + \cot^2 A) = \sin^2 A \cdot \csc^2 A = \sin^2 A \cdot \frac{1}{\sin^2 A} = 1\).

Question. If \(3 \sec \theta - 5 = 0\) then \(\cot \theta = ....................\) .
Answer: \(\frac{3}{4}\)
Explanation: \(3 \sec \theta - 5 = 0 \Rightarrow 3 \sec \theta = 5 \Rightarrow \sec \theta = \frac{5}{3}\).
\(\Rightarrow \sec^2 \theta = \frac{25}{9} \Rightarrow 1 + \tan^2 \theta = \frac{25}{9} \Rightarrow \tan^2 \theta = \frac{16}{9} \Rightarrow \tan \theta = \frac{4}{3} \Rightarrow \cot \theta = \frac{3}{4}\).

Question. If \(\sin \theta - \cos \theta = 0\), \(0 \leq \theta \leq 90^\circ\) then the value of \(\theta\) is .................... .
Answer: \(45^\circ\)
Explanation: \(\sin \theta - \cos \theta = 0 \Rightarrow \sin \theta = \cos \theta\). At \(\theta = 45^\circ\), \(\sin \theta = \frac{1}{\sqrt{2}}\) and \(\cos \theta = \frac{1}{\sqrt{2}}\). So, \(\theta = 45^\circ\).

Question. \(\cos 1^\circ \cos 2^\circ \cos 3^\circ ... \cos 180^\circ = .................... .\)
Answer: 0
Explanation: \(\cos 1^\circ \cos 2^\circ \cos 3^\circ ... \cos 180^\circ\). Since \(\cos 90^\circ = 0\), \(\therefore 0 \times \cos 1^\circ \cos 2^\circ .... \cos 180^\circ = 0\).

Question. If \(\tan \theta = \sqrt{3}\), then \(\sec \theta = ................... .\)
Answer: 2
Explanation: \(\tan \theta = \sqrt{3}\) gives \(\theta = 60^\circ\). So, \(\sec \theta = \sec 60^\circ = 2\).

Question. Maximum value of \(\frac{1}{\csc \theta}\) is .................... .
Answer: 1
Explanation: \(\frac{1}{\csc \theta} = \sin \theta\). As \(\sin \theta \leq 1\), thus maximum value of \(\frac{1}{\csc \theta}\) is 1.

Question. If \(\tan \theta + \cot \theta = 2\), then the value of \(\tan^2 \theta + \cot^2 \theta\) is .................... .
Answer: 2
Explanation: \(\tan^2 \theta + \cot^2 \theta = (\tan \theta + \cot \theta)^2 - 2 \tan \theta \cot \theta = (2)^2 - 2 \times 1 = 2\).

Very Short Answer Type Questions

Question. Write the value of \(\sin^2 30^\circ + \cos^2 60^\circ\). 
Answer: \(\frac{1}{2}\)
Explanation: \(= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)

Question. If \(\sin x + \cos y = 1\); \(x = 30^\circ\) and \(y\) is an acute angle, find the value of \(y\). 
Answer: Given, \(\sin x + \cos y = 1\) and \(x = 30^\circ\). \(\therefore \sin 30^\circ + \cos y = 1 \Rightarrow \frac{1}{2} + \cos y = 1 \Rightarrow \cos y = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow \cos y = \cos 60^\circ \therefore y = 60^\circ\). Hence, the value of \(y\) is \(60^\circ\).

Question. \(\sin^2 60^\circ + 2 \tan 45^\circ - \cos^2 30^\circ\)
Answer: 2
Explanation: \(\sin^2 60^\circ + 2 \tan 45^\circ - \cos^2 30^\circ\). Since, \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\tan 45^\circ = 1\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\). \(= \left(\frac{\sqrt{3}}{2}\right)^2 + 2 \times 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} + 2 - \frac{3}{4} = 2\).

Question. If \(\sin A = \frac{3}{4}\), calculate \(\sec A\). 
Answer: \(\frac{4}{\sqrt{7}}\)
Explanation: Given: \(\sin A = \frac{3}{4}\). Since, \(\sin^2 A + \cos^2 A = 1\), \(\cos^2 A = 1 - \sin^2 A\) or \(\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}\). \(\therefore \sec A = \frac{1}{\cos A} = \frac{4}{\sqrt{7}}\). Hence, the value of \(\sec A\) is \(\frac{4}{\sqrt{7}}\).

Question. If \((1 + \cos A) (1 - \cos A) = \frac{3}{4}\), find value of \(\sec A\). 
Answer: \(\pm 2\)
Explanation: \((1 + \cos A) (1 - \cos A) = \frac{3}{4} \Rightarrow 1 - \cos^2 A = \frac{3}{4} \Rightarrow \cos^2 A = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow \sec^2 A = 4 \Rightarrow \sec A = \pm 2\).

Question. The angle of elevation of the top of a tower at a point on the ground, 50 m away from the foot of the tower, is \(60^\circ\). Find the height of the tower.
Answer: \(50\sqrt{3}\) m
Explanation: Let, \(AB\) be the tower and C be the point of observation on the ground. From point C, the angle of elevation to the top of the tower is \(60^\circ\). In \(\Delta ABC\), \(\tan 60^\circ = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AB}{50} \Rightarrow AB = 50\sqrt{3} \text{ m}\). Hence, the height of the tower is \(50\sqrt{3} \text{ m}\).

Question. A ladder 15 m long makes an angle of \(60^\circ\) with the wall. Find the height of the point where the ladder touches the wall.
Answer: 7.5 m
Explanation: Let AC be the ladder of length 15 m, which is at the height \(AB\) i.e., ‘h’ m from the ground. The ladder makes an angle of \(60^\circ\) with the wall. Now, in \(\Delta ABC\), \(\cos 60^\circ = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{h}{15} \Rightarrow h = 7.5 \text{ m}\). Hence, the height of point where the ladder touches the wall is 7.5 m.

Question. If a tower 30 m high, casts a shadow \(10\sqrt{3}\) m long on the ground, then what is the angle of elevation of the sun? 
Answer: \(60^\circ\)
Explanation: Tower AB is 30m and shadow BC is \(10\sqrt{3}\) m. In \(\Delta ABC\) which is right triangle, \(\tan \theta = \frac{AB}{BC} = \frac{30}{10\sqrt{3}} \Rightarrow \tan \theta = \sqrt{3}\). But \(\tan 60^\circ = \sqrt{3}\). \(\therefore \theta = 60^\circ\). So, angle of elevation of sun is \(60^\circ\).

Question. If \(\tan A = 1\) and \(\cos B = \frac{1}{\sqrt{2}}\), then find \(\cos (A + B)\)
Answer: 0
Explanation: \(\tan A = 1 \Rightarrow A = 45^\circ\) and \(\cos B = \frac{1}{\sqrt{2}} \Rightarrow B = 45^\circ\). So, \(\cos (A + B) = \cos (45^\circ + 45^\circ) = \cos 90^\circ = 0\).

Question. Evaluate : \(2 \sec 30^\circ \times \tan 60^\circ\). 
Answer: 4
Explanation: \(2 \sec 30^\circ \times \tan 60^\circ = 2 \times \frac{2}{\sqrt{3}} \times \sqrt{3} = 4\).