If \(\sin A = \frac{12}{13}\), find the value of \(\sec A\).
Answer: B
Given \(\sin A = \frac{12}{13}\) (Opposite/Hypotenuse). We can find the adjacent side using the Pythagorean theorem: \(Adj = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\).
\(\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{13}\).
\(\sec A = \frac{1}{\cos A} = \frac{13}{5}\).
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What is the value of \(\tan 75° + \cot 75°\)?
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 \((2+\sqrt{3}) + (2-\sqrt{3}) = 4\).
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In a right-angled triangle ABC, right-angled at B, if \(\tan A = 1\), then the value of \(2 \sin A \cos A\) is:
Answer: B
If \(\tan A = 1\), then A = 45°.
The expression \(2 \sin A \cos A\) is the double angle identity for sine, \(\sin(2A)\).
So, we need to find \(\sin(2 \times 45°) = \sin(90°)\).
The value of \(\sin 90°\) is 1.
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The value of \(9 \sec^2 A - 9 \tan^2 A\) is:
Answer: B
We can factor out the common term, 9.
Expression = \(9 (\sec^2 A - \tan^2 A)\).
Using the Pythagorean identity \(\sec^2 A - \tan^2 A = 1\), the expression becomes:
9 × 1 = 9.
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From a point on the ground, 20 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?
Answer: B
Let h be the height and d be the distance from the foot.
\(\tan(\text{angle}) = \frac{\text{height}}{\text{distance}}\)
\(\tan 60° = \frac{h}{20}\)
\(\sqrt{3} = \frac{h}{20}\)
h = 20√3 meters.
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What is the value of \(\cos 38° \cos 52° - \sin 38° \sin 52°\)?
Answer: A
This expression is in the form of the cosine addition formula: \(\cos(A+B) = \cos A \cos B - \sin A \sin B\).
Here, A = 38° and B = 52°.
The expression is equal to \(\cos(38° + 52°) = \cos(90°)\).
The value of \(\cos 90°\) is 0.
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If \(\tan(A+B) = \sqrt{3}\) and \(\tan(A-B) = \frac{1}{\sqrt{3}}\), find A and B.
Answer: A
From the given information:
1) \(\tan(A+B) = \sqrt{3} \Rightarrow A+B = 60°\)
2) \(\tan(A-B) = \frac{1}{\sqrt{3}} \Rightarrow A-B = 30°\)
We have a system of two linear equations. Adding them gives:
2A = 90° => A = 45°.
Substituting A=45° into the first equation: 45° + B = 60° => B = 15°.
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If \(\cot \theta = 7/8\), the value of \(\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}\) is:
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 \((\frac{7}{8})^2 = \frac{49}{64}\).
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What is the value of \(\frac{2\tan 30°}{1 - \tan^2 30°}\)?
Answer: A
This expression is the double-angle identity for tangent: \(\tan(2A) = \frac{2\tan A}{1 - \tan^2 A}\).
Here, A = 30°.
The expression is equal to \(\tan(2 \times 30°) = \tan 60°\).
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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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