What is the minimum value of \(\sin^2\theta + \cos^2\theta\)?
Answer: B
This is a trick question based on the fundamental Pythagorean identity.
For any angle \(\theta\), the expression \(\sin^2\theta + \cos^2\theta\) is always equal to 1.
Therefore, its minimum value, maximum value, and average value are all 1.
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If a triangle ABC is right angled at C, then the value of \(\cos(A+B)\) is:
Answer: A
In a triangle the angles add to 180°, and \( C = 90^{\circ} \), so
\[ A + B = 90^{\circ} \]
\[ \cos(A+B) = \cos 90^{\circ} = 0 \]
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The value of \(\sin(2A)\) is given by:
Answer: C
The double-angle identity for sine is
\[ \sin 2A = 2\sin A \cos A \]
The option \( \cos^{2}A - \sin^{2}A \) is \( \cos 2A \), the companion identity - the pair most easily transposed.
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A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 40 m from the bank, he finds the angle to be 30°. The height of the tree is:
Answer: B
Let h be the height of the tree and x be the initial distance from the tree.
From the first observation: \(\tan 60° = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}\).
From the second observation: \(\tan 30° = \frac{h}{x+40} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x+40} \Rightarrow x+40 = h\sqrt{3}\).
Substitute h from the first equation into the second: \(x+40 = (x\sqrt{3})\sqrt{3} = 3x\).
\(40 = 2x \Rightarrow x = 20\) m.
Now find the height: \(h = x\sqrt{3} = 20\sqrt{3}\) m.
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If \(\sin A = 1/2\), then the value of \(\cot A\) is:
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 \(\cos 30° = \frac{\sqrt{3}}{2}\).
\(\cot 30° = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}\).
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The value of \((\sec A + \tan A)(1 - \sin A)\) is:
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:
\(\frac{\cos^2 A}{\cos A} = \cos A\).
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If \(\tan \theta = \frac{4}{3}\), then the value of \(\frac{3\sin \theta + 2\cos \theta}{3\sin \theta - 2\cos \theta}\) is:
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 \(\tan \theta = 4/3\):
\(\frac{3(4/3) + 2}{3(4/3) - 2} = \frac{4+2}{4-2} = \frac{6}{2} = 3\).
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The value of \(\sin 20° \cos 70° + \cos 20° \sin 70°\) is:
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.
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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 \(\cos 1° \cos 2° \cos 3° ... \cos 180°\) is:
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.
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