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
I. sin A + sin B = cos A + cos B
II. tan A + tan B = cot A + cot B
Select the correct answer using the codes given below.
(a) Only II
(b) Only I
(c) Both I and II
(d) Neither I nor II
The correct answers to the above question in:
Answer: (c)
Given, A = ${π}/6 \text"and B" = {π}/3$
I. L.H.S = sin A + sin B = sin ${π}/6 + sin {π}/3$
= $1/2 + {√3}/2 = {1 + √3}/2$
R.H.S = cos A + cos B = cos ${π}/6 + {cos π}/3$
${√3}/2 + 1/2 = {√3 + 1}/2$
⇒ sin A + sin B = cos A + cos B
II. L.H.S = tan A + tan B = tan ${π}/6 + tan {π}/3$
= $1/{√3} + √3 = 4/{√3}$
R.H.S = cot A + cot B = cot ${π}/6 + cot {π}/3$
= $√3 + 1/{√3} = 4/{√3}$
⇒ tan A + tan B = cot A + cot B
Both statements are true.
Alternate Method:
A + B = ${π}/6 + {π}/3 = {π}/2$
I. sin A + sin B = sin $({π}/2 - B) + sin ({π}/2 - A)$
= cos B + cos A = cos A + cos B
II. tan A + tan B = tan $({π}/2 - B) + tan ({π}/2 - A)$
= cot B + cot A = cot A + cot B
Hence, both statements are true.
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If 1 + tan θ = $√2$ , then what is the value of cot θ – 1?
a) $√2$
b) $1/{√2}$
c) 2
d) $1/2$
Answer »Answer: (a)
1 + tan θ = $√2$
⇒ tan θ = $√2$ - 1
∴ cot θ - 1 = $1/{√2 - 1} - 1 = {√2 + 1}/{2 - 1} - 1 = √2$
Question : 2
If $a^2 = {1 + 2 sin θ cos θ}/{1 - 2 sin θ cos θ}$, then what is the value of ${a + 1}/{a - 1}$ ?
a) 1
b) sec θ
c) 0
d) tan θ
Answer »Answer: (d)
Given that:
$a^2 = {\text"1 + 2 sin θ cos θ"}/{\text"1 - 2 sin θ cos θ"}$
⇒ $a^2 = {(sin^2 θ + cos^2 θ) + \text"2 sin θ . cos θ"}/{(sin^2 θ + cos^2 θ) - \text"2 sin θ . cos θ"}$
⇒ $a^2 = {(\text"sin θ + cos θ")^2}/{(\text"sin θ - cos θ")^2} ⇒ a/1 = {\text"sin θ + cos θ"}/{\text"sin θ - cos θ"}$
(applying componendo dividendo formula)
⇒ ${a + 1}/{a - 1} = {(\text"sin θ + cos θ") + (\text"sin θ - cos θ")}/{(\text"sin θ + cos θ") - (\text"sin θ - cos θ")}$
⇒ ${a + 1}/{a - 1} = {\text"2 sin θ"}/{\text"2 cos θ"}$ = tan θ
Question : 3
What is the value of tan1° tan2° tan3° tan4° ... tan89° ?
a) 1
b) 0
c) 2
d) $√3$
Answer »Answer: (a)
tan1° tan2° tan3° tan4°.......tan 89°
tan1° tan2° ......tan45°......tan 89°
= 1
Question : 4
ABC is a right triangle with right angle at A. If the value of tan B = $1/{√3}$, then for any real k the length of the hypotenuse is of the form
a) 2 k
b) 3 k
c) 5 k
d) 9 k
Answer »Answer: (a)
Given,
In ΔABC,
tan B = $k/{√3 k}$
By Pythogaros theorem,
$AB^2 + AC^2 = BC^2$
⇒ $(√3 k)^2 + (1k)^2 = BC^2$
⇒ $BC^2 = 4k^2$
⇒ BC = 2k
Question : 5
If tan (x + 40)° tan (x + 20)° tan (3x)° tan (70 – x)° tan (50 – x)° = 1, then the value of x is equal to
a) 20
b) 30
c) 15
d) 10
Answer »Answer: (c)
tan(x + 40)° tan (x + 20)° tan 3x° tan (70 - x)° tan (50 - x)° = 1
⇒ tan(x + 40) tan(x + 20) tan 3x cot [90 - (70 - x)] cot [90 - (50 - x)] = 1
⇒ tan(x + 40) tan (x + 20) tan 3x cot (x + 20) cot (x + 40) = 1
⇒ tan 3x = tan 45°
⇒ 3x = 45°
⇒ x = 15°
So, option (c) is correct.
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