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 \(\sin \theta = \frac{3}{5}\), what is the value of \(\cos \theta\)?
Answer: B
We use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\).
\((\frac{3}{5})^2 + \cos^2 \theta = 1\)
\(\frac{9}{25} + \cos^2 \theta = 1\)
\(\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}\)
\(\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}\) (Assuming θ is in the first quadrant).
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If \(\sin \theta + \sin^2 \theta = 1\), then the value of \(\cos^2 \theta + \cos^4 \theta\) is:
Answer: B
From the given equation, we can write:
\(\sin \theta = 1 - \sin^2 \theta\)
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we know that \(1 - \sin^2 \theta = \cos^2 \theta\).
Therefore, we have \(\sin \theta = \cos^2 \theta\).
Now we evaluate the expression \(\cos^2 \theta + \cos^4 \theta\). We can write \(\cos^4 \theta\) as \((\cos^2 \theta)^2\).
Substituting \(\cos^2 \theta = \sin \theta\), the expression becomes:
\(\sin \theta + (\sin \theta)^2 = \sin \theta + \sin^2 \theta\)
From the original given equation, we know that \(\sin \theta + \sin^2 \theta = 1\).
So, the value of the expression is 1.
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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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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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If \(\sin A + \csc A = 2\), then the value of \(\sin^2 A + \csc^2 A\) is:
Answer: B
Let \(x = \sin A\). Then \(\csc A = 1/x\). We are given \(x + 1/x = 2\). This is only possible if x=1. So, \(\sin A = 1\).
Then \(\csc A = 1/\sin A = 1\).
The expression becomes \(\sin^2 A + \csc^2 A = 1^2 + 1^2 = 1 + 1 = 2\).
Alternatively, square the given equation: \((\sin A + \csc A)^2 = 2^2\).
\(\sin^2 A + \csc^2 A + 2\sin A \csc A = 4\).
Since \(\sin A \csc A = 1\), we have \(\sin^2 A + \csc^2 A + 2 = 4\).
\(\sin^2 A + \csc^2 A = 2\).
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The value of \(\tan 315°\) is:
Answer: B
We can write 315° as 360° - 45°. The angle is in the fourth quadrant, where tangent is negative.
\(\tan(315°) = \tan(360° - 45°) = -\tan(45°)\).
Since \(\tan 45° = 1\), the value is -1.
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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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