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RRB JE Trigonometry Questions PDF

Download RRB JE Trigonometry  Questions and Answers PDF. Top 25 RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam

Question 1: Find the value of cos 1^\circ.cos 2^\circ.cos 3^\circ.cos 4^\circ……cos 179^\circ

a) -1

b) 0

c) $\sqrt{2}$

d) 1

Question 2: If sin x $= \dfrac{9}{41}$ (0 < x < $90^\circ$), then find the value of sec x – tan x.

a) $\dfrac{4}{5}$

b) $\dfrac{5}{4}$

c) $\dfrac{2}{5}$

d) $\dfrac{31}{40}$

Question 3: If sin x = \dfrac{7}{25}( 0 < x < 90^\circ), then find the value of tan x.

a) $\dfrac{24}{25}$

b) $\dfrac{7}{25}$

c) $\dfrac{14}{25}$

d) $\dfrac{15}{23}$

Question 4: A person observed that the angle of elevation to the top of tower from a certain distance from its foot is 45^\circ. He had moved 30 metres towards the tower and the angle of elevation is now increased to 60^\circ. Then find the height of the tower.

a) $45+10\sqrt{3}$

b) $35+15\sqrt{3}$

c) $45+15\sqrt{3}$

d) $50+10\sqrt{3}$

Question 5: What is the value of $\frac{\sin20\times\sin40\times\sin60}{\cos30\times\cos50\times\cos70}$

a) 1

b) 0

c) 1/4

d) 1/2

Question 6: Minimum value of $2\sin^{2}x+1\cos^{2}x$

a) 1.5

b) 2

c) 1

d) 0

Question 7: Maximum value of $12\sin^{2}x+13\cos^{2}x$

a) 24

b) 12

c) 1

d) 13

Question 8: $a\sec60$=$b\cos60$ and $c\sin30$=$d\csc60$,value of ac/bd is

a) $\frac{1}{\sqrt{3}}$

b) $\frac{2}{\sqrt{3}}$

c) $\frac{4}{\sqrt{3}}$

d) $\frac{8}{\sqrt{3}}$

Question 9: $a\cos60$=$b\tan60$ and $c\sec30$=$d\cot60$,value of ac/bd is

a) $\frac{4}{\sqrt{3}}$

b) $\frac{2}{\sqrt{3}}$

c) $\frac{8}{\sqrt{3}}$

d) $\frac{1}{\sqrt{3}}$

Question 10: $a\sin30$=$b\csc60$ and $c\tan45$=$d\sec60$,value of ac/bd is

a) $\frac{4}{\sqrt{3}}$

b) $\frac{2}{\sqrt{3}}$

c) $\frac{8}{\sqrt{3}}$

d) $\frac{16}{\sqrt{3}}$

Question 11: Find the value of $tan 48^\circ – tan 3^\circ – tan 48^\circ.tan 3^\circ$

a) 0

b) 1

c) -1

d) 2

Question 12: Find the value of $tan 1^\circ . tan 2^\circ . tan 3^\circ . . . . tan 89^\circ$

a) 0

b) -1

c) 1

d) $\sqrt{3}$

Question 13: A person observes the top of a tower at an angle of 45°. He walks for 20 m towards the tower and the angle of elevation changes to 60°. Find the height of the tower?

a) $30 + 10\sqrt{3}$m

b) $20 + 10\sqrt{5}$m

c) $30$m

d) $30 + 15\sqrt{3}$

Question 14: If cos x = $\dfrac{12}{13}$, then find the value of tan x.

a) $\dfrac{5}{13}$

b) $\dfrac{5}{12}$

c) $\dfrac{13}{12}$

d) $\dfrac{9}{13}$

Question 15: $1/sinA – SinA/Cos^2A$ = ?

a) $2Cot2A.secA$

b) $2Cos2A.cosecA$

c) $2Cos2A.secA$

d) $2Cot2A.cosecA$

Question 16: If sin x = $\dfrac{3}{4} ( 0 < \theta < 90^\circ )$, find the value of tan x.

a) $\dfrac{4}{5}$

b) $\dfrac{3}{4}$

c) $\dfrac{5}{4}$

d) $\dfrac{5}{3}$

Question 17: FInd secθ-tanθ, if secθ+tanθ = 1/3 ?

a) 0

b) 1

c) 2

d) 3

Question 18: Find cosecθ, if sin2θ=$\frac{\sqrt3}{2}$?

a) 0

b) 1

c) 2

d) 3

Question 19: What is the value of secθ+cosθ, if sinθ = 12/13 ?It is known that 0 < θ < 90

a) 194/65

b) 124/25

c) 169/25

d) 145/25

Question 20: In a cricket ground, a batsman hit the ball straight and it hit the roof of commentary box at a point P which is at a height of 49 meters from the ground level. Find the approximate distance between batsman and bowler, if the angles of elevation of P from the bowler and batsman are 35° and 30° respectively ? (take tan 35°=0.7)

a) 10 meters

b) 12 meters

c) 13 meters

d) 15 meters

Question 21: $sin^{-1}x + cos^{-1}x = ? (-1\leq x \leq 1)$

a) $\pi/2$

b) $\pi$

c) $\pi/3$

d) $2\pi$

Question 22: If tan(θ/2) = 5/12, then find sinθ = ?

a) 5/13

b) 119/120

c) 120/169

d) 12/13

Question 23: It is known that 0 < θ < 90 and it is given that cot θ = 5/12. Find the value of cosecθ

a) 5/13

b) 13/12

c) 12/5

d) 12/13

Question 24: If sec θ + tan θ = -2, then find sec θ – tan θ = ?

a) -1/2

b) 2

c) 1/2

d) -2

Question 25: Find tan θ, if secθ = 13/12 ?
It is known that 0 < θ < 90

a) 12/5

b) 5/12

c) 12/13

d) 5/13

cos 1^\circ.cos 2^\circ.cos 3^\circ.cos 4^\circ……cos 179^\circ
cos 1^\circ.cos 2^\circ.cos 3^\circ.cos 4^\circ…….cos 90^\circ…..cos 179^\circ
Since cos 90^\circ = 0, the given series multiplication equals to zero.

Given $sin x = \dfrac{9}{41}$
⇒ $sin^2 x = \dfrac{81}{1681}$

We know that $cos^2 x = 1-sin^2 x$
Then, $cos^2 x = 1-\dfrac{81}{1681}$
⇒ $cos^2 x = \dfrac{1600}{1681}$

⇒ $cos x = \dfrac{40}{41}$
Then, $sec x = \dfrac{41}{40}$

$tan x = \dfrac{sin x}{cos x}$

$= \drac{9}{40}$

$sec x – tan x = \dfrac{41}{40} – \drac{9}{40} = \dfrac{32}{40} = \dfrac{4}{5}$

Given $sin x = \dfrac{7}{25}$
⇒ $sin^2 x = \dfrac{49}{625}$
We know that $cos^2 x = 1-sin^2 x$
$cos^2 x = 1-\dfrac{49}{625}$
$= \dfrac{576}{625} = \dfrac{24}{25}$

$tan x = \dfrac{sin x}{cos x}$

$= \dfrac{(\dfrac{7}{25})}{\dfrac{24}{25}}$

$= \dfrac{7}{24}$

The below diagram represents the given situation.

Here, Height of the tower = y metres
From the figure,
$tan 45^\circ = \dfrac{y}{x+30}$
⇒ $1 = \dfrac{y}{x+30}$
⇒ $y = x+30$
⇒ $x = y-30$
$tan 60^\circ = \dfrac{y}{x}$
$\sqrt{3} = \dfrac{y}{x}$
Substituting x = y-30 in above equation
$\dfrac{y}{y-30} = \sqrt{3}$
$(y-30)\sqrt{3} = y$
$\sqrt{3}y-y = 30\sqrt{3}$
⇒ $y(\sqrt{3}-1) = 30\sqrt{3}$
⇒ $y = \dfrac{30\sqrt{3}}{\sqrt{3}-1}$
Rationalising above equation
$y = \dfrac{30\sqrt{3}}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

$y = \dfrac{30\sqrt{3}(\sqrt{3}+1}{2} = \dfrac{90+30\sqrt{3}}{2} = 45+15\sqrt{3} 5) Answer (A) We know that$\sin\theta$=$\cos(90-\theta)$So for$\sin20$we have$\cos70\sin40$we have$\cos50\sin60$we have$\cos30$Therefore require value is 1 6) Answer (C)$(\sin^{2}x+\cos^{2}x)+\sin^{2}x$=1+$\sin^{2}x$Minimum value of$\sin x$is when$x$=0 i.e$\sin 0$=0 Therefore 1+0=1 7) Answer (D) 12$(\sin^{2}x+\cos^{2}x)+\cos^{2}x$=12(1)+$\cos^{2}x$Maximum value of$\cos x$is when$x$=0 i.e$\cos x$=1 Therefore 12+1=13 8) Answer (A) we have a=($b\cos60*\cos60$) a=b($\frac{1}{4}$) (a/b)=$\frac{1}{4}(c/\sin30$)=$\frac{d}{\sin60}$(c/d)=$\frac{4}{\sqrt{3}}$(ac/bd)=$\frac{1}{\sqrt{3}}$9) Answer (D) we have (a/2)=($b\tan60$) (a/2)=$\frac{b}{\sqrt{3}}$(a/b)=$\frac{2}{\sqrt{3}}(c/\cos30$)=$\frac{d}{\sqrt{3}}\frac{2c}{\sqrt{3}}$=$\frac{d}{\sqrt{3}}$(c/d)=(1/2) (ac/bd)=$\frac{1}{\sqrt{3}}$10) Answer (C) we have (a/2)=(b/$\sin60$) (a/2)=$\frac{2b}{\sqrt{3}}$(a/b)=$\frac{4}{\sqrt{3}}$c=$\frac{d}{\cos60}$c=2d (c/d)=2 (ac/bd)=$\frac{8}{\sqrt{3}}$11) Answer (B)$48-3 = 45$Applying tan on both sides$tan (48-3)^\circ = tan 45^\circ$→ (1) We know that$tan (A-B)^\circ = \dfrac{tan A^\circ – tan B^\circ}{1+tan A^\circ.tan B^\circ}$Then, (1) becomes$\dfrac{tan 48^\circ – tan 3^\circ}{1+tan 48^\circ.tan 3^\circ} = 1$(Since,$tan 45^\circ = 1$)$tan 48^\circ – tan 3^\circ = 1+tan 48^\circ.tan 3^\circ$⇒$tan 48^\circ – tan 3^\circ – tan 48^\circ.tan 3^\circ = 1$12) Answer (C) The given equation can be written as$tan 1^\circ . tan 89^\circ . tan 2^\circ . tan 88^\circ . . . . tan 45^\circ= tan 1^\circ . tan (90-1)^\circ . tan 2^\circ . tan (90-2)^\circ . . . . tan 45^\circ$We know that$tan (90-x)^\circ = cot x^\circ$Then, the above equation becomes$tan 1^\circ . cot 1^\circ . tan 2^\circ . cot 2^\circ . . . . tan 45^\circ$Since tan x^\circ. cot x^\circ = 1, the above equation equals to 1. Hence,$tan 1^\circ . tan 2^\circ . tan 3^\circ . . . . tan 89^\circ = 1$13) Answer (A) The given situation can be represented through the following figure. We know that$Tan 45^\circ =$h/BC => 1 = h/BC => BC = h Now after walking for 20m, the angle changed to 60°. So we have Tan 60 = h/BD =>$ \sqrt{3} = \frac{h}{h – 20}$=>$20\sqrt{3} = h(\sqrt{3} – 1)$=> h =$\frac{20\sqrt{3}}{\sqrt{3} – 1}$=> h =$ \frac{60 + 20\sqrt{3}}{2}$=> h =$ 30 + 10\sqrt{3}$14) Answer (B) Given cos x =$\dfrac{12}{13}$Squaring on both sides$cos^{2}x = \dfrac{144}{169}$We know that$sin^2 x = 1-cos^2 x$⇒$sin^2 x = 1-$\dfrac{144}{169}$

⇒ $sin^2 x =$\dfrac{25}{169}$⇒$sin x = $\dfrac{5}{13}$

We know that tan x = $\dfrac{sin x}{cos x}$

tan x $= \dfrac(\dfrac{5}{13})}{(\dfrac{12}{13})} = \dfrac{5}{12}$

$1/sinA – SinA/Cos^2A$
= $(cos^2A-sin^2A)/(sinA.cos^2A)$
= $2(Cos2A)/(sin2A.CosA)$
= $2Cot2A.secA$
So the answer is option A.

We know that cos x = $\sqrt{1-sin^{2}x}$
$= \sqrt{1-\dfrac{9}{25}}$
$= \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$
tan x = \dfrac{sin x}{cos x} = \dfrac{(\dfrac{3}{5})}{(\dfrac{4}{5})} = \dfrac{3}{4}$17) Answer (D) W.K.T$sec^2\theta – tan^2\theta = 1(sec\theta-tan\theta)(sec\theta+tan\theta) = 1(sec\theta-tan\theta)(1/3) = 1sec\theta-tan\theta = 3$So the answer is option D. 18) Answer (C) sin2θ=$\frac{\sqrt3}{2}$then 2θ = 60° ==> θ = 30° cosecθ = cosec30° = 2 So the answer is option C. 19) Answer (A) If sinθ = 12/13 cosθ = 5/13 and secθ = 13/5 secθ+cosθ = 13/5 + 5/13 = 194/65 So the answer is option A. 20) Answer (D) Tan35° = 49/y 0.7 = 49/y y = 70 Tan30° = 49/(x+y) 1/√3 = 49/(x+70) x+70 = 49√3 x+70 = 84.87 x = 84.87-70 = 14.87$\sim$15 meters So the answer is option D. 21) Answer (A) For any value of x,$sin^{-1}x + cos^{-1}x = \pi/2$Lets take x = 1/2$sin^{-1}(1/2) + cos^{-1}(1/2) = 30 + 60 = 90$°=$\pi/2$So the answer is option A. 22) Answer (C)$Sinθ$=$\frac{2tan(θ/2)}{1+tan^2(θ/2)}$=$\frac{2(5/12)}{1+(5/12)^2}$=$120/169$So the answer is option C. 23) Answer (B)$\text{cot} \theta = \frac{5}{12}\text{cosec}^2 \theta$=$1+\text{cot}^2 \theta = 1+{(\frac{5}{12})}^2 = \frac{169}{144}\text{cosec}\theta = \frac{13}{12}$So the answer is option B. 24) Answer (A)$sec^2 θ – tan^2 θ = 1(sec θ – tan θ)(sec θ + tan θ) = 1(sec θ – tan θ) (-2) = 1(sec θ – tan θ) = -1/2$So the answer is option A. 25) Answer (B)$sec^2θ-tan^2θ = 1(13/12)^2-tan^2θ = 1169/144-tan^2θ = 1tan^2θ = 25/144tanθ = 5/12\$