Competitive Exams : Quantitative Aptitude - Trig (Set 5) | Concept of Maths

Question 1

1. If $\sin \theta + \sin^2 \theta = 1$, then find the value of $\cos^{12} \theta + 3\cos^{10} \theta + 3\cos^8 \theta + \cos^6 \theta - 1$.

A) 1

B) -1

C) 0

D) 2

Answer

Correct Answer: C

Given $\sin \theta = 1 - \sin^2 \theta = \cos^2 \theta$. Substituting $\cos^2 \theta = \sin \theta$ into the expression: $(\cos^4 \theta + \cos^2 \theta)^3 - 1$ $= ((\cos^2 \theta)^2 + \cos^2 \theta)^3 - 1$ $= (\sin^2 \theta + \sin \theta)^3 - 1$ Since $\sin \theta + \sin^2 \theta = 1$, we get: $(1)^3 - 1 = 1 - 1 = 0$.

Question 2

2. A man observes the angle of elevation of the top of a tower to be $45^\circ$. Walking $30$ meters towards the tower, the angle becomes $60^\circ$. What is the height of the tower (in meters)?

A) $15(\sqrt{3}-1)$

B) $15(3+\sqrt{3})$

C) $30(\sqrt{3}+1)$

D) $45$

Answer

Correct Answer: B

Let $h$ be the height. In the first case: $\tan 45^\circ = \frac{h}{d} \Rightarrow h = d$. In the second case: $\tan 60^\circ = \frac{h}{d-30} \Rightarrow \sqrt{3} = \frac{h}{h-30}$. $h\sqrt{3} - 30\sqrt{3} = h \Rightarrow h(\sqrt{3}-1) = 30\sqrt{3}$. $h = \frac{30\sqrt{3}}{\sqrt{3}-1} = \frac{30\sqrt{3}(\sqrt{3}+1)}{2} = 15(3+\sqrt{3})$.

Question 3

3. What is the minimum value of $16 \sec^2 \theta + 25 \csc^2 \theta$ for $0 < \theta < \frac{\pi}{2}$?

A) 81

B) 41

C) 9

D) 40

Question 4

4. If $x = a \cos^3 \alpha$ and $y = b \sin^3 \alpha$, then find the value of $(\frac{x}{a})^{2/3} + (\frac{y}{b})^{2/3}$.

A) $a^2+b^2$

B) $0$

C) $\frac{1}{ab}$

D) 1

Answer

Correct Answer: D

$(\frac{x}{a}) = \cos^3 \alpha \Rightarrow (\frac{x}{a})^{2/3} = (\cos^3 \alpha)^{2/3} = \cos^2 \alpha$. Similarly, $(\frac{y}{b})^{2/3} = (\sin^3 \alpha)^{2/3} = \sin^2 \alpha$. Sum $= \cos^2 \alpha + \sin^2 \alpha = 1$.

Question 5

5. Find the value of $\tan 1^\circ \tan 2^\circ \tan 3^\circ ... \tan 89^\circ$.

A) 1

B) 0

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

D) $\infty$

Answer

Correct Answer: A

Pairs like $\tan 1^\circ \cdot \tan 89^\circ$ can be written as $\tan 1^\circ \cdot \cot 1^\circ = 1$ (since $\tan(90-\theta) = \cot \theta$). There are 44 such pairs and the middle term $\tan 45^\circ = 1$. Result $= 1 \times 1 \times ... \times 1 = 1$.

Question 6

6. If $\sec \theta + \tan \theta = p$, then find the value of $\sin \theta$.

A) $\frac{p^2+1}{p^2-1}$

B) $\frac{p^2-1}{2p}$

C) $\frac{p^2-1}{p^2+1}$

D) $\frac{2p}{p^2+1}$

Answer

Correct Answer: C

Since $\sec^2 \theta - \tan^2 \theta = 1$, we have $(\sec \theta - \tan \theta) = 1/p$. Adding equations: $2\sec \theta = p + 1/p = \frac{p^2+1}{p}$. Subtracting: $2\tan \theta = p - 1/p = \frac{p^2-1}{p}$. $\sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{p^2-1}{p^2+1}$.

Question 7

7. Find the maximum value of $5 \sin \theta + 12 \cos \theta + 7$.

A) 13

B) 20

C) 19

D) 25

Answer

Correct Answer: B

The maximum value of $a \sin \theta + b \cos \theta$ is $\sqrt{a^2 + b^2}$. Max value of $5 \sin \theta + 12 \cos \theta = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$. Total Max value $= 13 + 7 = 20$.

Question 8

8. If $x \cos A - y \sin A = 1$ and $x \sin A + y \cos A = 4$, find the value of $x^2 + y^2$.

A) 5

B) 16

C) 15

D) 17

Answer

Correct Answer: D

Squaring and adding both equations: $(x \cos A - y \sin A)^2 + (x \sin A + y \cos A)^2 = 1^2 + 4^2$ $x^2 \cos^2 A + y^2 \sin^2 A - 2xy \sin A \cos A + x^2 \sin^2 A + y^2 \cos^2 A + 2xy \sin A \cos A = 17$ $x^2(\cos^2 A + \sin^2 A) + y^2(\sin^2 A + \cos^2 A) = 17$ $x^2 + y^2 = 17$.

Question 9

9. If $\tan (A+B) = \sqrt{3}$ and $\tan (A-B) = 1$, find the value of $A$ (where $A, B$ are acute).

A) $45^\circ$

B) $52.5^\circ$

C) $60^\circ$

D) $37.5^\circ$

Answer

Correct Answer: B

$A+B = 60^\circ$ (since $\tan 60^\circ = \sqrt{3}$) $A-B = 45^\circ$ (since $\tan 45^\circ = 1$) Adding: $2A = 105^\circ \Rightarrow A = 52.5^\circ$.

Question 10

10. Simplify: $\frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta}$.

A) $2\tan \theta$

B) $2\sin \theta$

C) $2\sec \theta$

D) $\cos^2 \theta$

Answer

Correct Answer: C

$\frac{\cos \theta(1+\sin \theta) + \cos \theta(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$ $= \frac{\cos \theta + \sin \theta \cos \theta + \cos \theta - \sin \theta \cos \theta}{1 - \sin^2 \theta}$ $= \frac{2\cos \theta}{\cos^2 \theta} = \frac{2}{\cos \theta} = 2\sec \theta$.

Competitive Exams : Quantitative Aptitude - Trig (Set 5) | Concept of Maths