Trigonometry Questions and Answers

Answer: A

We substitute the standard trigonometric values:

\(\tan 45° = 1\)

\(\cos 60° = 1/2\)

\(\sin 60° = \sqrt{3}/2\)

\(\cot 60° = 1/\sqrt{3}\)

The equation becomes: \(x \cdot 1 \cdot (1/2) = (\sqrt{3}/2) \cdot (1/\sqrt{3})\).

\(x/2 = 1/2 \Rightarrow x = 1\).

Answer: B

We can rewrite the expression using the identity \(\sin^2\theta + \cos^2\theta = 1\).

\(2\sin^2\theta + 3\cos^2\theta = 2\sin^2\theta + 2\cos^2\theta + \cos^2\theta\)

= \(2(\sin^2\theta + \cos^2\theta) + \cos^2\theta\)

= \(2(1) + \cos^2\theta = 2 + \cos^2\theta\).

The value of \(\cos^2\theta\) ranges from 0 to 1. To get the minimum value of the expression, we take the minimum value of \(\cos^2\theta\), which is 0.

Minimum value = 2 + 0 = 2.

Answer: B

Convert to sines and cosines:

\((\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2 = (\frac{1 - \cos \theta}{\sin \theta})^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}\).

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)\), the expression becomes:

\(\frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1-\cos \theta}{1+\cos \theta}\).

Answer: B

We can write 15° as the difference of two standard angles, 45° - 30°.

Using the sine subtraction formula: \(\sin(A-B) = \sin A \cos B - \cos A \sin B\).

\(\sin(45°-30°) = \sin 45° \cos 30° - \cos 45° \sin 30°\)

= \((\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) - (\frac{1}{\sqrt{2}})(\frac{1}{2})\)

= \(\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}\).

Answer: B

Use the complementary-angle identity \( \cos(90^{\circ} - \theta) = \sin\theta \).

\[ \cos 72^{\circ} = \cos(90^{\circ} - 18^{\circ}) = \sin 18^{\circ} \]

\[ \frac{\sin 18^{\circ}}{\sin 18^{\circ}} = 1 \]

Neither value has to be computed - only recognised as equal.

Answer: B

Let the height be h and the shadow length be s. We are given h = s.

The angle of elevation θ is given by \(\tan \theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{s}\).

Since h = s, \(\tan \theta = 1\).

The angle whose tangent is 1 is 45°.

Answer: C

We use the identity \(\cot \theta = \tan(90° - \theta)\).

So, \(\cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A)\).

The equation becomes \(\tan 2A = \tan(108° - A)\).

This implies \(2A = 108° - A\).

3A = 108°

A = 36°.

Answer: B

Let H be the height of the tower and h=25m be the height of the cliff. Let d be the horizontal distance between them. Let θ be the angle.

Angle of depression to the foot: \(\tan \theta = \frac{h}{d} = \frac{25}{d}\).

Angle of elevation to the top: \(\tan \theta = \frac{H-h}{d} = \frac{H-25}{d}\).

Since the angles are equal, their tangents are equal.

\(\frac{25}{d} = \frac{H-25}{d} \Rightarrow 25 = H - 25 \Rightarrow H = 50\) m.

Answer: B

This is a fundamental Pythagorean identity, derived by dividing \(\sin^2 A + \cos^2 A = 1\) by \(\sin^2 A\).

\(\frac{\sin^2 A}{\sin^2 A} + \frac{\cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A}\)

\(1 + \cot^2 A = \csc^2 A\).

Answer: B

Correct answer: 25/3

Factor the 3 out and use a Pythagorean identity rather than finding \(\tan X\) first.

\[3 + 3\tan^2 X = 3\left(1 + \tan^2 X\right) = 3\sec^2 X\]

Since \(\cos X = \tfrac{3}{5}\), we have \(\sec X = \tfrac{5}{3}\), so

\[3\sec^2 X = 3 \times \frac{25}{9} = \frac{25}{3}\]

The same result follows the long way: \(\cos X = \tfrac{3}{5}\) gives a 3–4–5 triangle with \(\tan X = \tfrac{4}{3}\), and \(3 + 3\left(\tfrac{16}{9}\right) = 3 + \tfrac{16}{3} = \tfrac{25}{3}\). Stopping at \(\tfrac{16}{3}\) omits the leading 3.