Trigonometry Questions and Answers

Answer: B

The signs of the ratios identify the quadrant.

• Quadrant I: both positive
• Quadrant II: \( \sin > 0 \), \( \cos < 0 \) ✓
• Quadrant III: both negative
• Quadrant IV: \( \sin < 0 \), \( \cos > 0 \)

So \( \theta \) lies in the second quadrant. The mnemonic "all, sine, tangent, cosine" names the ratio that is positive in each quadrant in turn.

Answer: A

We use the product-to-sum formula \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\).

Here, A = 75° and B = 15°.

The expression equals \(\cos(75°+15°) + \cos(75°-15°) = \cos 90° + \cos 60°\).

Substituting values: 0 + 1/2 = 1/2.

Answer: A

The third angle is 180° - 30° - 45° = 105°. Let the included side be 'c'. We can use the formula Area = \(\frac{c^2 \sin A \sin B}{2 \sin C}\), where C is the angle opposite side c. Here C=105°.

Area = \(\frac{(\sqrt{3}+1)^2 \sin 30° \sin 45°}{2 \sin 105°}\). This is too complex. Let's try another way. Drop a perpendicular from the vertex with angle 105 to the side c. Let it be h. Let the side c be split into x and y. Then \(\tan 30 = h/x\) and \(\tan 45 = h/y\). So \(x=h\sqrt{3}\) and \(y=h\). We have \(x+y = h(\sqrt{3}+1) = \sqrt{3}+1\). This means h=1. Area = \(1/2 \times base \times height = 1/2 \times (\sqrt{3}+1) \times 1 = \frac{\sqrt{3}+1}{2}\) cm².

Answer: C

Find a common denominator: \(\tan A (\sec A + 1)\).

Expression = \(\frac{\tan^2 A + (\sec A + 1)^2}{\tan A (\sec A + 1)}\).

Numerator = \(\tan^2 A + \sec^2 A + 2\sec A + 1\). Using \(\tan^2 A = \sec^2 A - 1\), it becomes \((\sec^2 A - 1) + \sec^2 A + 2\sec A + 1 = 2\sec^2 A + 2\sec A = 2\sec A(\sec A + 1)\).

So, Expression = \(\frac{2\sec A(\sec A + 1)}{\tan A (\sec A + 1)} = \frac{2\sec A}{\tan A} = \frac{2/\cos A}{\sin A/\cos A} = \frac{2}{\sin A} = 2\csc A\).

Answer: B

Let h be the height and x be the final distance from the lighthouse.

From the second observation: \(\tan 45° = h/x \Rightarrow 1 = h/x \Rightarrow h = x\).

From the first observation, the distance was x+100: \(\tan 30° = \frac{h}{x+100} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{h+100}\) (since x=h).

\(h+100 = h\sqrt{3} \Rightarrow 100 = h(\sqrt{3}-1) \Rightarrow h = \frac{100}{\sqrt{3}-1} = \frac{100(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{100(\sqrt{3}+1)}{3-1} = 50(\sqrt{3}+1)\) m.

Answer: B

The expression simplifies using the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\).

Numerator = \(1 - \sin^2\theta = \cos^2\theta\).

Denominator = \(1 - \cos^2\theta = \sin^2\theta\).

The expression becomes \(\frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta\).

Given \(\cot \theta = 7/8\), the value is \((\frac{7}{8})^2 = \frac{49}{64}\).

Answer: B

We use the complementary angle identities: \(\sec(90°-A) = \csc A\) and \(\cot(90°-A) = \tan A\).

\(\sec(15°-\theta) = \sec(90° - (75°+\theta)) = \csc(75°+\theta)\).

\(\cot(35°-\theta) = \cot(90° - (55°+\theta)) = \tan(55°+\theta)\).

The expression becomes: \(\csc(75°+\theta) - \csc(75°+\theta) - \tan(55°+\theta) + \tan(55°+\theta) = 0\).

Answer: C

Subtract a full turn to bring the angle into range.

\[ 420^{\circ} - 360^{\circ} = 60^{\circ} \]

\[ \sin 60^{\circ} = \frac{\sqrt{3}}{2} \]

Sine has period 360°, so the value repeats exactly.

Answer: D

If \(\tan \theta = 12/5\) (Opposite/Adjacent), we find the hypotenuse: \(Hyp = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169} = 13\).

\(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13}\).

The expression is \(1 + \frac{12}{13} = \frac{13+12}{13} = \frac{25}{13}\).

Answer: C

Convert to sin and cos: \(\frac{\sin A}{1-\frac{\cos A}{\sin A}} + \frac{\cos A}{1-\frac{\sin A}{\cos A}} = \frac{\sin^2 A}{\sin A - \cos A} + \frac{\cos^2 A}{\cos A - \sin A}\)

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

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

= \(\frac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A - \cos A} = \sin A + \cos A\).