Trigonometric Ratios & Identity Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Trigonometric Ratios & Identities topic of quantitative aptitude
(a) positive but greater than 1
(b) positive but less than 1
(c) negative
(d) 0
The correct answers to the above question in:
Answer: (b)
Since, value of cos θ decreases, from 0° to 90° and at 45°, it is equal to the value of sin θ.
Similarly, value of sin θ increases from 0° to 90° and at 45°, it is equal to the value of cos θ
For 0° < 45°, cos θ > sin θ
So, value of cos 25° is always positive but less than 1.
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If the given figure, BC = 15 cm and sin B = $4/5$. What is the value of AB?
a) 20 cm
b) 25 cm
c) 5 cm
d) 4 cm
Answer »Answer: (b)
BC = 15 cm and sin B = $4/5$
sin B = ${AC}/{AB} = 4/5$
then BC = 3m
But, BC = 15 (given)
then AC = 4 × 5 = 20
AB = 5 × 5 = 25.
Hence, the value of AB is 25 cm.
Question : 2
If tan θ + sec θ = 2, then tan θ is equal to
a) $5/4$
b) $3/4$
c) $3/2$
d) $5/2$
Answer »Answer: (b)
tan θ + sec θ = 2 ......(i)
As we know
⇒ $sec^2 θ - tan^2 θ$ = 1
⇒ (sec θ - tan θ)(sec θ + tan θ) = 1
⇒ sec θ - tan θ = $1/2$ ....(ii)
equation (i) - eq (ii)-
2 tan θ = 2 - 1/2
2 tan θ = $3/2 ⇒ tan θ = 3/4$
So, option (b) is correct.
Question : 3
If sec x cosec x = 2, then what is $tan^n x + cot^n$ x equal to?
a) $2^{n + 1}$
b) 2
c) 2n
d) $2^{n – 1}$
Answer »Answer: (b)
sec x cosec x = 2
This value is possible is x = 45°
$tan^n x + cot^n x = (\text"tan x")^n + (\text"cot x")^n$
= $(tan 45°)^n + (cot 45°)^n = (1)^n + (1)^n = 2$
Question : 4
If 0 ≤ θ ≤ ${π}/2$ and cos θ + $√3$ sin θ = 2, then what is the value of θ ?
a) ${π}/4$
b) ${π}/3$
c) ${π}/6$
d) ${π}/2$
Answer »Answer: (b)
Given that, cos θ + $√3$ sin θ = 2
⇒ $1/2 cos θ + {√3}/2$ sin θ = 1
⇒ sin 30° cos θ + cos 30° sin θ = 1
⇒ sin(30° + θ) = sin 90°
30° + θ = 90°
∴ θ = 60°
Question : 5
If sin θ + cos θ = 1, then what is the value of sin θ cos θ ?
a) 0
b) 2
c) 1
d) $1/2$
Answer »Answer: (a)
Given, $\text"sin θ + cos θ"$ = 1
Squaring both sides,
$(sin^2 θ + cos^2 θ) + 2 \text"sin θ cos θ" = 1$
⇒ 1 + 2 sin θ cos θ = 1 ⇒ sin θ cos θ = 0.
Question : 6
If cos A = tan B, cos B = tan C and cos C = tan A then sin A is equal to
a) ${√5 - 1}/2$
b) ${√5 - 1}/4$
c) ${√3 - 1}/4$
d) None of these
Answer »Answer: (d)
Cos A = tan B
Squaring on both sides
$cos^2 A = tan^2 B$
⇒ $tan^2 B = {sin^2 B}/{cos^2 B} = {1 - cos^2 B}/{cos^2 B}$
∴ $cos^2 A = {1 - cos^2 B}/{cos^2 B}$
$cos^2 A = {1 - tan^2 C}/{tan^2 C}$
(∵ cos B = tan C)
⇒ $cos^2 A tan^2 C = 1 - tan^2 C = 1 - {sin^2 C}/{cos^2 C} $
⇒ $cos^2 A {sin^2 C}/{cos^2 C} ={cos^2 C - sin^2 C}/{cos^2 C}$
⇒ $cos^2 A (1 - cos^2 C) = 2 cos^2 C - 1$
⇒ $cos^2 A (1 - tan^2 A) = 2 tan^2 A - 1$
$cos^2 A - sin^2 A = {2 sin^2 A}/{cos^2 A} - 1$
⇒ 1 - $2sin^2 A = {2 sin^2 A - cos^2 A}/{cos^2 A}$
⇒ $cos^2 A (1 - sin^2 A) = 2 sin^2 A - cos^2 A$
⇒ $cos^2 A (1 - 2 sin^2 A) = 2 sin^2 A - 1 + sin^2 A$
⇒ $(1 - sin^2 A) (1 - 2 sin^2 A) = 3 sin^2 A - 1$
⇒ $1 - 2 sin^2 A - sin^2 A + 2 sin^4A = 3 sin^2 A - 1$
⇒ $1 - 3 sin^2 A + 2 sin^4 A = 3 sin^2 A - 1$
⇒ $2 sin^4 A - 6 sin^2 A + 2 = 0$
⇒ $sin^4 A - 3 sin^2$ A + 1 = 0
This is quadratic equation in $sin^2$ A
$(sin^2 A)^2 - 3 (sin^2 A) + 1 = 0$
$sin^2 A = {3 ± √{(- 3)^2 - 4 (1) (1)}}/2$
= ${3 ± √5}/2$
$sin^2 A = {3 + √5}/2$ not possible because $sin^2 A$ ≯ 1
So $sin^2 A = {3 - √5}/2$
So none of the options are correct.
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