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).
Enter details here
If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is:
Answer: B
Let h be the height of the tower and s be the length of the shadow. The angle of elevation θ is given by \(\tan \theta = h/s\).
Since the height h is constant, if the shadow length 's' increases, the value of the fraction h/s decreases.
As \(\tan \theta\) decreases, the angle θ also decreases (for acute angles). Therefore, the angle of elevation is decreasing.
Enter details here
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 \(\frac{\sin 18°}{\sin 18°} = 1\).
Enter details here
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 \((2+\sqrt{3}) + (2-\sqrt{3}) = 4\).
Enter details here
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 \(\tan \theta = 4/3\):
\(\frac{3(4/3) + 2}{3(4/3) - 2} = \frac{4+2}{4-2} = \frac{6}{2} = 3\).
Enter details here
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.
Enter details here
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 \(\tan(2 \times 30°) = \tan 60°\).
Enter details here
The value of \(\frac{1-\tan^2 45°}{1+\tan^2 45°}\) is:
Answer: D
We know that \(\tan 45° = 1\).
Substitute this value into the expression:
\(\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\).
Alternatively, this is the formula for \(\cos(2\theta)\), so it equals \(\cos(2 \times 45°) = \cos 90° = 0\).
Enter details here
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, \(\sec \theta = \frac{x^2+1}{2x}\).
Enter details here
If the angles of a triangle are 30° and 45° and the included side is \((\sqrt{3}+1)\) cm, then the area of the triangle is:
Answer: A
The third angle is 180° - 30° - 45° = 105°. Let the included side be 'c'. We can use the formula Area = \(\frac{c^2 \sin A \sin B}{2 \sin C}\), where C is the angle opposite side c. Here C=105°.
Area = \(\frac{(\sqrt{3}+1)^2 \sin 30° \sin 45°}{2 \sin 105°}\). This is too complex. Let's try another way. Drop a perpendicular from the vertex with angle 105 to the side c. Let it be h. Let the side c be split into x and y. Then \(\tan 30 = h/x\) and \(\tan 45 = h/y\). So \(x=h\sqrt{3}\) and \(y=h\). We have \(x+y = h(\sqrt{3}+1) = \sqrt{3}+1\). This means h=1. Area = \(1/2 \times base \times height = 1/2 \times (\sqrt{3}+1) \times 1 = \frac{\sqrt{3}+1}{2}\) cm².
Enter details here