If \(A+B=90°\), then \(\sqrt{\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}}\) is equal to:
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 \(\sqrt{\tan^2 A} = \tan A\).
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The value of \(\tan 1° \tan 2° \tan 3° ... \tan 89°\) is:
Answer: B
We use the complementary angle identity \(\tan(90° - \theta) = \cot \theta\), and the fact that \(\tan \theta \cdot \cot \theta = 1\).
The expression can be paired up: \((\tan 1° \tan 89°) (\tan 2° \tan 88°) ...\)
Since \(\tan 89° = \tan(90°-1°) = \cot 1°\), the first pair is \(\tan 1° \cot 1° = 1\).
All pairs will multiply to 1. The middle term is \(\tan 45°\), which is also 1.
The final product is 1.
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What is the maximum value of \(3\sin \theta + 4\cos \theta\)?
Answer: C
For an expression of the form \(a\sin \theta + b\cos \theta\), the maximum value is \(\sqrt{a^2+b^2}\) and the minimum value is \(-\sqrt{a^2+b^2}\).
Here, a = 3 and b = 4.
Maximum value = \(\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\).
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If \(\sec \theta + \tan \theta = x\), then \(\sec \theta\) is equal to:
Answer: A
We know that \(\sec^2 \theta - \tan^2 \theta = 1\), which can be factored as \((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\).
Given \(\sec \theta + \tan \theta = x\), we have \(\sec \theta - \tan \theta = \frac{1}{x}\).
Now we have a system of two linear equations:
1) \(\sec \theta + \tan \theta = x\)
2) \(\sec \theta - \tan \theta = 1/x\)
Adding the two equations gives: \(2\sec \theta = x + \frac{1}{x} = \frac{x^2+1}{x}\).
Therefore, \(\sec \theta = \frac{x^2+1}{2x}\).
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If a pole 6 m high casts a shadow 2√3 m long on the ground, then the sun's angle of elevation is:
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°.
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The value of \(\frac{\sin 18°}{\cos 72°}\) is:
Answer: B
We use the complementary angle identity \(\cos(90° - \theta) = \sin \theta\).
So, \(\cos 72° = \cos(90° - 18°) = \sin 18°\).
The expression becomes \(\frac{\sin 18°}{\sin 18°} = 1\).
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A 20 m long ladder is placed against a wall such that it reaches a height of 10 m. What is the angle the ladder makes with the ground?
Answer: A
The ladder is the hypotenuse (20 m) and the height on the wall is the opposite side (10 m).
Let θ be the angle with the ground.
\(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{10}{20} = \frac{1}{2}\).
The angle whose sine is 1/2 is 30°.
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If \(\tan \theta = \frac{a}{b}\), then \(b\cos 2\theta + a\sin 2\theta\) is equal to:
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 \(b(\frac{b^2-a^2}{a^2+b^2}) + a(\frac{2ab}{a^2+b^2}) = \frac{b^3-a^2b+2a^2b}{a^2+b^2} = \frac{b^3+a^2b}{a^2+b^2} = \frac{b(b^2+a^2)}{a^2+b^2} = b\).
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A ramp for a wheelchair has an angle of elevation of 5°. If the ramp is 12 feet long, how high off the ground is the top of the ramp? (Given sin 5° ≈ 0.087)
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.
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The expression \(\sin^2 A + \cos^2 A\) is equal to:
Answer: B
This is the most fundamental Pythagorean identity in trigonometry. For any angle A, the sum of the square of its sine and the square of its cosine is always equal to 1.
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