Trigonometry Questions and Answers

Answer: A

If \(\sin A = 1/2\), then A = 30°.

We need to find the value of \(\cot 30°\).

\(\cot A = \frac{\cos A}{\sin A}\). We know \(\cos 30° = \frac{\sqrt{3}}{2}\).

\(\cot 30° = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}\).

Answer: B

Let's convert everything to sines and cosines.

\((\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A)\)

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

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

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

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

\(\frac{\cos^2 A}{\cos A} = \cos A\).

Answer: B

We can divide both the numerator and the denominator by \(\cos \theta\).

Expression = \(\frac{3(\frac{\sin \theta}{\cos \theta}) + 2(\frac{\cos \theta}{\cos \theta})}{3(\frac{\sin \theta}{\cos \theta}) - 2(\frac{\cos \theta}{\cos \theta})} = \frac{3\tan \theta + 2}{3\tan \theta - 2}\).

Substitute \(\tan \theta = 4/3\):

\(\frac{3(4/3) + 2}{3(4/3) - 2} = \frac{4+2}{4-2} = \frac{6}{2} = 3\).

Answer: B

This expression is in the form of the sine addition formula: \(\sin(A+B) = \sin A \cos B + \cos A \sin B\).

Here, A = 20° and B = 70°.

The expression is equal to \(\sin(20° + 70°) = \sin(90°)\).

The value of \(\sin 90°\) is 1.

Answer: C

Let the angle of elevation be θ.

\(\tan \theta = \frac{\text{Height of pole}}{\text{Length of shadow}} = \frac{6}{2\sqrt{3}}\).

\(\tan \theta = \frac{3}{\sqrt{3}} = \sqrt{3}\).

The angle whose tangent is √3 is 60°.

Answer: B

The product runs through every whole degree from 1 to 180, and one of those factors is

\[ \cos 90^{\circ} = 0 \]

A single zero factor makes the whole product 0, whatever the other 179 terms are.

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: \(\sin(A-B) = \sin A \cos B - \cos A \sin B\).

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

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

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