Trigonometry Questions and Answers

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

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

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

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

Answer: B

Let h be the height and L be the length of the ramp.

\(\sin(\text{angle}) = \frac{\text{height}}{\text{length}}\)

\(\sin 5° = \frac{h}{12}\)

h = 12 × \(\sin 5°\) ≈ 12 × 0.087 = 1.044 feet.

Answer: A

Since A+B=90°, we have B=90-A. So, \(\tan B = \cot A\), \(\cot B = \tan A\), \(\sec B = \csc A\), \(\sin B = \cos A\).

Substitute these into the expression under the root:

Numerator: \(\tan A \cot A + \tan A \tan A = 1 + \tan^2 A = \sec^2 A\).

Denominator: \(\sin A \csc A = 1\).

Second term: \(\frac{\cos^2 A}{\cos^2 A} = 1\).

The expression becomes \(\sqrt{\frac{\sec^2 A}{1} - 1} = \sqrt{\sec^2 A - 1}\).

Using the identity \(\sec^2 A - 1 = \tan^2 A\), we get \(\sqrt{\tan^2 A} = \tan A\).

Answer: B

315° lies in the fourth quadrant, where the tangent is negative.

\[ \tan 315^{\circ} = -\tan(360^{\circ} - 315^{\circ}) = -\tan 45^{\circ} = -1 \]

Answer: B

Sine is an odd function.

\[ \sin(-\theta) = -\sin\theta \]

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

Cosine, by contrast, is even: \( \cos(-\theta) = \cos\theta \).

Answer: C

We rewrite the expression using identities:

\(\sec^2\theta + \csc^2\theta = (1+\tan^2\theta) + (1+\cot^2\theta) = 2 + \tan^2\theta + \cot^2\theta\).

Let \(y = \tan^2\theta\). The expression is \(2 + y + 1/y\). By AM-GM inequality, \(y+1/y \ge 2\sqrt{y \cdot 1/y} = 2\). The minimum value of \(y+1/y\) is 2.

Therefore, the minimum value of the entire expression is 2 + 2 = 4.

Answer: D

Let the two legs of the right triangle be a and b. Hypotenuse c = 25 cm.

Perimeter = a + b + c = 60. So, a + b + 25 = 60 => a + b = 35.

By Pythagorean theorem, a² + b² = c² = 25² = 625.

We know \((a+b)^2 = a^2+b^2+2ab\). So, \(35^2 = 625 + 2ab \Rightarrow 1225 = 625 + 2ab \Rightarrow 2ab = 600 \Rightarrow ab = 300\).

We need two numbers that add to 35 and multiply to 300. The numbers are 15 and 20.

The legs are 15 cm and 20 cm. Let one acute angle be A. The tangent values would be opposite/adjacent, which could be 15/20 = 3/4 or 20/15 = 4/3.