Competitive Exams : Quantitative Aptitude - Trig (Set 4) | Concept of Maths

Question 1

1. Find the minimum value of the expression $16\tan^2\theta + 9\cot^2\theta + 7$.

A) 31

B) 25

C) 32

D) 24

Answer

Correct Answer: A

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality: For positive terms $a$ and $b$, $a + b \ge 2\sqrt{ab}$. Let $a = 16\tan^2\theta$ and $b = 9\cot^2\theta$. $$16\tan^2\theta + 9\cot^2\theta \ge 2\sqrt{16\tan^2\theta \cdot 9\cot^2\theta}$$ Since $\tan^2\theta \cdot \cot^2\theta = 1$: $$16\tan^2\theta + 9\cot^2\theta \ge 2\sqrt{144} = 2 \cdot 12 = 24$$ Minimum value of the whole expression $= 24 + 7 = 31$.

Question 2

2. If $\tan\theta + \cot\theta = 2$, then find the value of $\sin^{2026}\theta + \cos^{2026}\theta$.

A) 2

B) 1

C) $2^{-1012}$

D) $2^{1013}$

Answer

Correct Answer: C

Given $\tan\theta + \frac{1}{\tan\theta} = 2$. This is in the form $x + \frac{1}{x} = 2$, which implies $x = 1$. So, $\tan\theta = 1$, which means $\theta = 45^\circ$ (for the first quadrant). Now, $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$. $$\left(\frac{1}{\sqrt{2}}\right)^{2026} + \left(\frac{1}{\sqrt{2}}\right)^{2026} = 2 \cdot \left(\frac{1}{2}\right)^{1013} = 2^1 \cdot 2^{-1013} = 2^{-1012}$$

Question 3

3. If $A + B = 45^\circ$, what is the value of $(1 + \tan A)(1 + \tan B)$?

A) 1

B) 2

C) 0

D) 4

Question 4

4. How many solutions does the equation $\sin^2\theta - 3\sin\theta + 2 = 0$ have in the interval $[0, 2\pi]$?

A) 1

B) 2

C) 0

D) Infinite

Answer

Correct Answer: A

Let $y = \sin\theta$. The quadratic equation is $y^2 - 3y + 2 = 0$. $(y - 1)(y - 2) = 0 \implies y = 1$ or $y = 2$. 1) If $\sin\theta = 1$, then $\theta = \frac{\pi}{2}$ (One solution in $[0, 2\pi]$). 2) If $\sin\theta = 2$, this is impossible since the range of $\sin\theta$ is $[-1, 1]$. Total number of solutions = 1.

Question 5

5. Simplify: $\cos 20^\circ \cos 40^\circ \cos 80^\circ$

A) 1/2

B) 1/4

C) 1/6

D) 1/8

Answer

Correct Answer: D

Using the identity: $\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta \dots \cos(2^{n-1}\theta) = \frac{\sin(2^n\theta)}{2^n \sin \theta}$ Here $\theta = 20^\circ$ and $n = 3$. $$\frac{\sin(2^3 \cdot 20^\circ)}{2^3 \sin 20^\circ} = \frac{\sin 160^\circ}{8 \sin 20^\circ}$$ Since $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$: $$\frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}$$

Question 6

6. If $\sin A + \sin B = a$ and $\cos A + \cos B = b$, find the value of $\tan\left(\frac{A + B}{2}\right)$.

A) $b/a$

B) $a/b$

C) $a+b$

D) $\sqrt{a^2+b^2}$

Answer

Correct Answer: B

Using sum-to-product identities: $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) = a$ ... (1) $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) = b$ ... (2) Dividing (1) by (2): $$\frac{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}{2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)} = \frac{a}{b}$$ $$\tan\left(\frac{A+B}{2}\right) = \frac{a}{b}$$

Question 7

7. A vertical pole $5$ m high stands on top of a building $10$ m high. To an observer at a point $10$ m away from the base of the building, the pole subtends an angle $\theta$. Find $\tan \theta$.

A) 1/3

B) 1/2

C) 1/5

D) 2/5

Answer

Correct Answer: C

Let $\alpha$ be the angle of elevation of the top of the building. $\tan \alpha = \frac{10}{10} = 1$. Let $\beta$ be the angle of elevation of the top of the pole. The total height is $10 + 5 = 15$ m. $\tan \beta = \frac{15}{10} = 1.5$. The angle subtended by the pole is $\theta = \beta - \alpha$. $$\tan \theta = \tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha}$$ $$\tan \theta = \frac{1.5 - 1}{1 + (1.5)(1)} = \frac{0.5}{2.5} = \frac{1}{5}$$

Question 8

8. Find the value of: $\log(\tan 1^\circ) + \log(\tan 2^\circ) + \dots + \log(\tan 89^\circ)$

A) 0

B) 1

C) 1/2

D) Not defined

Answer

Correct Answer: A

Using the property of logarithms: $\log a + \log b = \log(ab)$. Sum $= \log(\tan 1^\circ \cdot \tan 2^\circ \cdot \dots \cdot \tan 89^\circ)$ Since $\tan(90^\circ - \theta) = \cot \theta$: $\tan 89^\circ = \cot 1^\circ$, $\tan 88^\circ = \cot 2^\circ$, and so on. $\tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \cot 1^\circ = 1$. The product inside the log is 1 (with $\tan 45^\circ = 1$ in the middle). Sum $= \log(1) = 0$.

Question 9

9. What is the range of the function $f(x) = 3\sin x + 4\cos x + 10$?

A) $[3, 17]$

B) $[5, 15]$

C) $[7, 13]$

D) $[0, 20]$

Answer

Correct Answer: B

The range of $a\sin x + b\cos x$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$. Here $a=3, b=4$. Range of $3\sin x + 4\cos x$ is $[-\sqrt{3^2+4^2}, \sqrt{3^2+4^2}] = [-5, 5]$. Adding 10 to all parts: Minimum value $= -5 + 10 = 5$. Maximum value $= 5 + 10 = 15$. Range $= [5, 15]$.

Question 10

10. If $\sin\theta - \cos\theta = \frac{7}{13}$ and $0 < \theta < 90^\circ$, find the value of $\sin\theta + \cos\theta$.

A) 13/17

B) 1

C) 15/13

D) 17/13

Answer

Correct Answer: D

Let $x = \sin\theta + \cos\theta$. We know the identity: $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2(\sin^2\theta + \cos^2\theta)$ $x^2 + \left(\frac{7}{13}\right)^2 = 2(1)$ $x^2 + \frac{49}{169} = 2$ $x^2 = 2 - \frac{49}{169} = \frac{338 - 49}{169} = \frac{289}{169}$ $x = \sqrt{\frac{289}{169}} = \frac{17}{13}$ (taking positive since $\theta$ is in the first quadrant).

Competitive Exams : Quantitative Aptitude - Trig (Set 4) | Concept of Maths