Trigonometry Questions and Answers

Answer: C

First, find the values of \(\tan 75°\) and \(\cot 75°\).

\(\tan 75° = \tan(45°+30°) = \frac{\tan 45 + \tan 30}{1 - \tan 45 \tan 30} = \frac{1+1/\sqrt{3}}{1-1/\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}\).

\(\cot 75° = 1/\tan 75° = \frac{1}{2+\sqrt{3}} = 2-\sqrt{3}\).

The sum is \((2+\sqrt{3}) + (2-\sqrt{3}) = 4\).

Answer: B

This is the double-angle formula in reverse.

\[ 2\sin A\cos A = \sin 2A \]

\[ 2\sin 15^{\circ}\cos 15^{\circ} = \sin 30^{\circ} = \frac{1}{2} \]

Recognising the form avoids having to evaluate either 15° value.

Answer: B

We use the double angle formulas in terms of tan θ:

\(\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}\) and \(\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}\)

Substitute \(\tan \theta = a/b\):

\(\cos 2\theta = \frac{1-a^2/b^2}{1+a^2/b^2} = \frac{b^2-a^2}{b^2+a^2}\)

\(\sin 2\theta = \frac{2(a/b)}{1+a^2/b^2} = \frac{2ab}{a^2+b^2}\)

The expression is \(b(\frac{b^2-a^2}{a^2+b^2}) + a(\frac{2ab}{a^2+b^2}) = \frac{b^3-a^2b+2a^2b}{a^2+b^2} = \frac{b^3+a^2b}{a^2+b^2} = \frac{b(b^2+a^2)}{a^2+b^2} = b\).

Answer: A

When the hands coincide they point in exactly the same direction, so the angle between them is 0°.

They are 180° apart when opposite and 90° apart at a right angle - the other two configurations these questions ask about.

Answer: C

This is in the form of the cosine double angle identity: \(\cos(2A) = 1 - 2\sin^2 A\).

Here, A = 22.5°.

The expression is equal to \(\cos(2 \times 22.5°) = \cos(45°)\).

The value of \(\cos 45°\) is \(1/\sqrt{2}\).

Answer: B

We are given \(\tan \theta + \cot \theta = 2\). Squaring both sides:

\((\tan \theta + \cot \theta)^2 = 2^2\)

\(\tan^2 \theta + \cot^2 \theta + 2 \tan \theta \cot \theta = 4\)

Since \(\tan \theta \cot \theta = 1\), the equation becomes:

\(\tan^2 \theta + \cot^2 \theta + 2(1) = 4\)

\(\tan^2 \theta + \cot^2 \theta = 2\).

Answer: B

We use the complementary angle identity \(\sin(90° - \theta) = \cos \theta\).

\(\sin 65° = \sin(90° - 25°) = \cos 25°\).

The expression becomes \(\sin^2 25° + \cos^2 25°\).

Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), the value is 1.

Answer: A

Let L be the length of the string (hypotenuse) and h be the height (opposite side).

\(\sin 60° = \frac{\text{height}}{\text{length}} = \frac{75}{L}\)

\(\frac{\sqrt{3}}{2} = \frac{75}{L}\)

\(L = \frac{75 \times 2}{\sqrt{3}} = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3}\) m.

Answer: A

This is the double-angle formula for the tangent.

\[ \tan 2A = \frac{2\tan A}{1 - \tan^{2}A} \]

With \( A = 30^{\circ} \):

\[ \frac{2\tan 30^{\circ}}{1 - \tan^{2}30^{\circ}} = \tan 60^{\circ} \]

Checking numerically: \( \tfrac{2/\sqrt{3}}{1 - 1/3} = \tfrac{2/\sqrt{3}}{2/3} = \sqrt{3} = \tan 60^{\circ} \).

Answer: B

Given \(\sin A = \frac{12}{13}\) (Opposite/Hypotenuse). We can find the adjacent side using the Pythagorean theorem: \(Adj = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\).

\(\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{13}\).

\(\sec A = \frac{1}{\cos A} = \frac{13}{5}\).