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# RRB NTPC Mathematics Questions: 29th June

Dear aspirants,

As you all know, RRB NTPC Exam will be held in the month of July-September 2019. So we are here to help you with the Mathematics subject. We are providing daily quantitative aptitude quizzes, which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.

Q1. If (tan⁡θ+cot⁡θ)/(tan⁡θ-cot⁡θ ) = 2, (0 ≤ θ ≤ 90°), then the value of sin θ is
यदि (tan⁡θ+cot⁡θ)/(tan⁡θ-cot⁡θ ) = 2, (0 ≤ θ ≤ 90°), तो sin θ का मान ज्ञात कीजिये
(a) 2/√3
(b) √3/2
(c) 1/2
(d) 1
Ans.(b)
Sol. Applying C & D
(2 tan⁡θ)/(2 cot⁡θ )=3/1
sin^2⁡θ/cos^2⁡θ =3/1
sin² θ = 3(1 – sin² θ)
4sin² θ = 3
sin θ = √3/2

(a) –1
(b) 0
(c) 1
(d) 2
Ans.(c)
Sol.
⇒(2cos⁡22°)/cos⁡22°-(2tan⁡75°)/(5tan⁡75°)-(3tan⁡45°.tan⁡(90-70°).tan⁡(90 -50°).tan⁡50°.tan⁡70°)/5
=2-2/5-3/5
= 2 – 1 =1

Q3. If cos^4⁡θ- sin^4⁡θ=2/3, then the value of 2 cos² θ – 1 is
यदि cos^4⁡θ- sin^4⁡θ=2/3 है,तो 2 cos² θ – 1 का मान ज्ञात कीजिये
(a) 0
(b) 1
(c) 2/3
(d) 3/2
Ans.(c)
Sol. (sin² θ + cos² θ) (cos² θ – sin² θ) = 2/3
2 cos² θ – 1 = 2/3

Q4. If sin α sec (30° + α) = 1 (0 < α < 60°), then the value of sin α + cos 2α is
यदि sin α sec (30° + α) = 1 (0 < α < 60°) है,तो sin α + cos 2α का मान कितना है
(a) 1
(b) (2+√3)/(2√3)
(c) 0
(d) √2
Ans.(a)
Sol. sin⁡α/cos⁡(30°+α) =1
sin⁡α/sin⁡(90-30-α) =1
sin⁡α/sin⁡(60-α) =1
sin α = sin (60 – α)
α = 60 – α
α = 30°
sin 30° + cos 60°
=1/2+1/2=1

Q5. If (sin⁡θ+cos⁡θ)/(sin⁡θ-cos⁡θ ) = 3, then the value of sin^4⁡θ-cos^4⁡θ is
यदि (sin⁡θ+cos⁡θ)/(sin⁡θ-cos⁡θ ) = 3 है,तो sin^4⁡θ-cos^4⁡θ का मान कितना है
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Ans.(c)
Sol. (sin⁡θ+cos⁡θ)/(sin⁡θ-cos⁡θ )=3
sin θ + cos θ = 3sin θ – 3cos θ
2sin θ = 4cos θ
tan θ = 2
sin^4⁡θ-cos^4⁡θ
= (sin² θ + cos² θ) (sin² θ – cos² θ)
= sin² θ – cos² θ
= cos² θ (tan² θ – 1)
=(tan^2⁡θ-1)/sec^2⁡θ =(tan^2⁡θ-1)/(1+tan^2⁡θ )= (4-1)/(1+4)=3/5

Q6. If sec² θ + tan² θ = 7, then the value of θ when 0° ≤ θ ≤ 90°, is
यदि sec² θ + tan² θ = 7 है,तो θ का मान ज्ञात कीजिये जब 0° ≤ θ ≤ 90° है
(a) 60°
(b) 30°
(c) 0°
(d) 90°
Ans.(a)
Sol. sec² θ + tan² θ = 7
1 + tan² θ + tan² θ = 7
2tan² θ = 6
tan² θ = 3
tan θ = √3
θ = 60°

Q7. The simplified value of (sec x sec y + tan x tan y)² – (sec x tan y + tan x sec y)² is:
(sec x sec y + tan x tan y)² – (sec x tan y + tan x sec y)² का सरलीकृत मूल्य है:
(a) –1
(b) 0
(c) sec² x
(d) 1
Ans.(d)
Sol.
=sec² x sec² y + tan² x tan² y + 2(sec x sec y) (tan x tan y) – sec² x tan² y – tan² x sec² y – 2(sec x tan y) (tan x sec y)
= sec² x sec² y + tan² x. tan² y – sec² x tan² y – tan² x sec² y
= sec² x sec² y – sec² x. tan² y – tan² x sec² y + tan² x tan² y
= sec² x (sec² y – tan² y) – tan² x (sec² y – tan² y)
= (sec² y – tan² y) (sec² x – tan² x)
= 1 × 1 = 1

Q8. If sin θ + cosec θ = 2, then value of sin^100⁡θ+ cosec^100 θ is equal to :
यदि sin θ + cosec θ = 2 है,तो sin^100⁡θ+ cosec^100 θ का मान किसके बराबर है:
(a) 1
(b) 2
(c) 3
(d) 100
Ans.(b)
Sol. sin θ + 1/sin⁡θ = 2
sin² θ – 2sin θ + 1 = 0
(sin θ – 1) = 0
sin θ = 1
sin^100⁡θ+cosec^100 θ = 1 + 1 = 2

Q9. If cos² α + cos² β = 2, then the value of tan³ α + sin^5⁡β is :
यदि cos² α + cos² β = 2 है,तो tan³ α + sin^5⁡β का मान है:
(a) –1
(b) 0
(c) 1
(d) 1/√3
Ans.(b)
Sol. cos² α + cos² β = 2
1 – sin² α + 1 – sin² β = 2
sin² α + sin² β = 0
sin² α = 0 & sin² β = 0
α = 0 & β = 0
tan³ α + sin^5⁡β = 0

Q10. If θ is a positive acute angle and tan 2θ tan 3θ = 1, then the value of (2 cos^2⁡(5θ/2)-1)
यदि θ एक सकारात्मक न्यून कोण है और tan 2θ tan 3θ = 1 है,तो (2 cos^2⁡(5θ/2)-1) का मान ज्ञात कीजिये
(a) -1/2
(b) 1
(c) 0
(d) 1/2
Ans.(c)
Sol. tan 2θ.tan 3θ = 1
tan⁡3θ=1/tan⁡2θ
tan 3θ = cot 2θ
tan 3θ = tan (90 – 2θ)
3θ = 90 – 2θ
5θ = 90
θ = 18°
2 cos^2(⁡5θ/2)-1=2 cos^2⁡(45)-1
=2×1/2-1
= 1 – 1 = 0

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