Competitive Exams : Quantitative Aptitude - Trig (Set 6) | Concept of Maths

Question 1

1. Find the maximum value of the expression: $3 \sin x + 4 \cos x + 12$

A) 12

B) 19

C) 17

D) 5

Answer

Correct Answer: C

The maximum value of $a \sin x + b \cos x$ is given by $\sqrt{a^2 + b^2}$. Here $a=3, b=4$. Maximum of $3 \sin x + 4 \cos x = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$. Thus, the maximum value of the entire expression is $5 + 12 = 17$.

Question 2

2. If $\sec \theta + \tan \theta = 4$, what is the value of $\cos \theta$?

A) $\frac{15}{17}$

B) $\frac{8}{17}$

C) $\frac{4}{17}$

D) $\frac{1}{4}$

Answer

Correct Answer: B

We know that $\sec^2 \theta - \tan^2 \theta = 1$. $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$. Given $\sec \theta + \tan \theta = 4$, then $\sec \theta - \tan \theta = \frac{1}{4}$. Adding the two equations: $2 \sec \theta = 4 + \frac{1}{4} = \frac{17}{4}$. $\sec \theta = \frac{17}{8} \implies \cos \theta = \frac{8}{17}$.

Question 3

3. Evaluate the sum: $\sin^2 1^\circ + \sin^2 2^\circ + \sin^2 3^\circ + \dots + \sin^2 89^\circ$

A) 44.5

B) 45

C) 89

D) 44

Answer

Correct Answer: A

Note that $\sin^2 \theta + \sin^2(90^\circ - \theta) = \sin^2 \theta + \cos^2 \theta = 1$. Pairs $(1, 89), (2, 88), \dots, (44, 46)$ each sum to 1. There are 44 such pairs. The middle term is $\sin^2 45^\circ = (\frac{1}{\sqrt{2}})^2 = 0.5$. Total Sum = $44 \times 1 + 0.5 = 44.5$.

Question 4

Question 5

5. What is the value of $\tan 15^\circ + \cot 15^\circ$?

A) 2

B) 4

C) $2\sqrt{3}$

D) 1

Answer

Correct Answer: B

$\tan 15^\circ + \cot 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} + \frac{\cos 15^\circ}{\sin 15^\circ} = \frac{\sin^2 15^\circ + \cos^2 15^\circ}{\sin 15^\circ \cos 15^\circ} = \frac{1}{\sin 15^\circ \cos 15^\circ}$. Multiply and divide by 2: $\frac{2}{2 \sin 15^\circ \cos 15^\circ} = \frac{2}{\sin 30^\circ} = \frac{2}{1/2} = 4$.

Question 6

6. Find the minimum value of $9 \sec^2 \theta + 4 \csc^2 \theta$ for $0 < \theta < \frac{\pi}{2}$.

A) 13

B) 12

C) 25

D) 36

Answer

Correct Answer: C

The minimum value of $a^2 \sec^2 \theta + b^2 \csc^2 \theta$ is $(a + b)^2$. Here $a^2 = 9 \implies a=3$ and $b^2 = 4 \implies b=2$. Min value = $(3 + 2)^2 = 5^2 = 25$.

Question 7

7. If $\sin \theta + \csc \theta = 2$, then the value of $\sin^{50} \theta + \csc^{50} \theta$ is:

A) 2

B) 1

C) 50

D) $2^{50}$

Answer

Correct Answer: A

$\sin \theta + \frac{1}{\sin \theta} = 2 \implies \sin^2 \theta - 2 \sin \theta + 1 = 0 \implies (\sin \theta - 1)^2 = 0$. So, $\sin \theta = 1$. Consequently, $\csc \theta = 1$. Therefore, $1^{50} + 1^{50} = 1 + 1 = 2$.

Question 8

8. Simplify: $(\sec A - \cos A)(\csc A - \sin A)(\tan A + \cot A)$

A) 0

B) $\sin A \cos A$

C) $\sec A \csc A$

D) 1

Answer

Correct Answer: D

$(\sec A - \cos A) = \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A}$. $(\csc A - \sin A) = \frac{1 - \sin^2 A}{\sin A} = \frac{\cos^2 A}{\sin A}$. $(\tan A + \cot A) = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$. Multiplying: $\frac{\sin^2 A}{\cos A} \times \frac{\cos^2 A}{\sin A} \times \frac{1}{\sin A \cos A} = 1$.

Question 9

9. If $7 \sin^2 \theta + 3 \cos^2 \theta = 4$, then what is the value of $\tan \theta$ (assuming acute $\theta$)?

A) $\frac{1}{\sqrt{3}}$

B) $\sqrt{3}$

C) 1

D) 0

Answer

Correct Answer: A

$7 \sin^2 \theta + 3 (1 - \sin^2 \theta) = 4 \implies 7 \sin^2 \theta + 3 - 3 \sin^2 \theta = 4$. $4 \sin^2 \theta = 1 \implies \sin^2 \theta = \frac{1}{4} \implies \sin \theta = \frac{1}{2}$. For acute angle, $\theta = 30^\circ$. $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Question 10

10. A ladder 20m long just reaches the top of a wall. If the ladder makes an angle of $30^\circ$ with the wall, find the height of the wall.

A) 10m

B) 20m

C) $10\sqrt{3}$m

D) 15m

Answer

Correct Answer: C

The angle with the wall is $30^\circ$. Therefore, the angle with the ground is $90^\circ - 30^\circ = 60^\circ$. Let height be $H$. $\sin 60^\circ = \frac{H}{\text{hypotenuse}} = \frac{H}{20}$. $H = 20 \times \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$m.

Accepted Answer

Correct Answer: D

$x^2 + y^2 = r^2 \sin^2 A \cos^2 B + r^2 \sin^2 A \sin^2 B = r^2 \sin^2 A (\cos^2 B + \sin^2 B) = r^2 \sin^2 A$. Now, $(x^2 + y^2) + z^2 = r^2 \sin^2 A + r^2 \cos^2 A = r^2(\sin^2 A + \cos^2 A) = r^2$.

Competitive Exams : Quantitative Aptitude - Trig (Set 6) | Concept of Maths