trigonometric ratios and identities mcq Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024
ibps rrb so officer sclae 2 3 single exam 2024 SYLLABUS WISE SUBJECTS MCQs
Number System
Lcm & Hcf
Ratio & Proportion
Percentage
Time & Work
Mensuration
Advance Math
Trigonometric Ratios & Identities
Average
Profit & Loss
If cot θ (1 + sin θ) = 4m and cot θ (1 – sin θ) = 4n, then which of the following is correct?
Answer: (a)
Put θ = 45°, we get
m = ${√2 + 1}/{4 √2} \text"and n" = {√2 - 1}/{4 √2}$
R.H.S:
mn= ${(√2 + 1)}/{4 √ 2} {√2 - 1}/{4 √2} = {2 - 1}/{32} = 1/{32}$
L.H.S: $(m^2 - n^2)^2$
$[({√2 + 1}/{4 √2})^2 - ({√2 - 1}/{4 √2})^2]^2 = [{3 + 2 √2}/{32} - {(3 - 2 √2)}/{32}]^2$
$({3 + 2 √2 - 3 + 2 √2}/{32})^{32} = ({4 √2}/{32})^2 = ({√2}/8)^2$
= $(1/{4 √2})^2 = 1/{32}$
If 0 < x < 45° and 45° < y < 90°, then which one of the following is correct?
Answer: (a)
As we know, sin x is increasing from 0 to 90°
∴ sin y > sin x.
If cot θ = ${2xy}/{x^2 - y^2}$, then what is cos θ equal to?
Answer: (c)
Given, cot θ = ${2xy}/{x^2 - y^2}$
In ΔABC,
$AC^2 = (x^2 - y^2)^2 + (2xy)^2$
⇒ $AC^2 = (x^2 + y^2)^2 ⇒ AC= x^2 + y^2$
∴ cos θ = ${AB}/{AC} = {2xy}/{x^2 + y^2}$
If sin x + cos x = p and $sin^3 x + cos^3$ x = q, then what is $p^3$ – 3p equal to ?
Answer: (a)
Let sin x + cos x = p .....(i)
$sin^3 + cos^3x$ = q ......(ii)
On cubing Eq. (i) both sides
$sin^3 x + cos^3 \text"x + 3 sin x cos x (sin x + cos x)" = p^3$
Put $sin^3 x + cos^3 x$ = q from equation (ii)
⇒ q + 3 sin x cos x(p) = $p^3$ ....(iii)
On squaring Eq. (i) both sides, we get
$sin^2 x + cos^2 \text"x + 2 sin x cos x" = p^2$
⇒ sin x cos x = ${p^2 - 1}/2$ [∵ $sin^2 x + cos^2 x = 1$]
From Eq. (iii),
q + ${3 (p^2 - 1)}/2 = p^3$
⇒ 2q + $3p^3 - 3p = 2p^3$ ⇒ $p^3$ - 3p = - 2q
What is ${sin^6 θ - cos^6 θ}/{sin^2 θ - cos^2 θ}$ equal to?
Answer: (a)
${sin^6 θ - cos^6 θ}/{sin^2 θ - cos^2 θ} = {(sin^2 θ)^3 - (cos^2 θ)^3}/{sin^2 θ - cos^2 θ}$
= ${(sin^2 θ - cos^2 θ) (sin^4 θ + cos^4 θ + sin^2 θ cos^2 θ)}/{sin^2 θ - cos^2 θ}$
= $sin^4 θ + cos^4 θ + 2 sin^2 θ cos^2 θ - sin^2 θ cos^2 θ$
= $(sin^2 θ + cos^2 θ)^2 - sin^2 θ cos^2 θ = 1 - sin^2 θ cos^2 θ$
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