{"id":26094,"date":"2019-03-15T17:16:53","date_gmt":"2019-03-15T11:46:53","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26094"},"modified":"2019-03-15T17:16:53","modified_gmt":"2019-03-15T11:46:53","slug":"rrb-je-trigonometry-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/rrb-je-trigonometry-questions-pdf\/","title":{"rendered":"RRB JE Trigonometry Questions PDF"},"content":{"rendered":"

RRB JE Trigonometry Questions PDF<\/span><\/h1>\n

Download RRB JE Trigonometry\u00a0 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<\/p>\n

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Question 1:\u00a0<\/b>Find the value of cos 1^\\circ.cos 2^\\circ.cos 3^\\circ.cos 4^\\circ\u2026…cos 179^\\circ<\/p>\n

a)\u00a0-1<\/p>\n

b)\u00a00<\/p>\n

c)\u00a0$\\sqrt{2}$<\/p>\n

d)\u00a01<\/p>\n

Question 2:\u00a0<\/b>If sin x $= \\dfrac{9}{41}$ (0 < x < $90^\\circ$), then find the value of sec x – tan x.<\/p>\n

a)\u00a0$\\dfrac{4}{5}$<\/p>\n

b)\u00a0$\\dfrac{5}{4}$<\/p>\n

c)\u00a0$\\dfrac{2}{5}$<\/p>\n

d)\u00a0$\\dfrac{31}{40}$<\/p>\n

Question 3:\u00a0<\/b>If sin x = \\dfrac{7}{25}( 0 < x < 90^\\circ), then find the value of tan x.<\/p>\n

a)\u00a0$\\dfrac{24}{25}$<\/p>\n

b)\u00a0$\\dfrac{7}{25}$<\/p>\n

c)\u00a0$\\dfrac{14}{25}$<\/p>\n

d)\u00a0$\\dfrac{15}{23}$<\/p>\n

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Question 4:\u00a0<\/b>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.<\/p>\n

a)\u00a0$45+10\\sqrt{3}$<\/p>\n

b)\u00a0$35+15\\sqrt{3}$<\/p>\n

c)\u00a0$45+15\\sqrt{3}$<\/p>\n

d)\u00a0$50+10\\sqrt{3}$<\/p>\n

Question 5:\u00a0<\/b>What is the value of $\\frac{\\sin20\\times\\sin40\\times\\sin60}{\\cos30\\times\\cos50\\times\\cos70}$<\/p>\n

a)\u00a01<\/p>\n

b)\u00a00<\/p>\n

c)\u00a01\/4<\/p>\n

d)\u00a01\/2<\/p>\n

Question 6:\u00a0<\/b>Minimum value of $2\\sin^{2}x+1\\cos^{2}x$<\/p>\n

a)\u00a01.5<\/p>\n

b)\u00a02<\/p>\n

c)\u00a01<\/p>\n

d)\u00a00<\/p>\n

Question 7:\u00a0<\/b>Maximum value of $12\\sin^{2}x+13\\cos^{2}x$<\/p>\n

a)\u00a024<\/p>\n

b)\u00a012<\/p>\n

c)\u00a01<\/p>\n

d)\u00a013<\/p>\n

Question 8:\u00a0<\/b>$a\\sec60$=$b\\cos60$ and $c\\sin30$=$d\\csc60$,value of ac\/bd is<\/p>\n

a)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n

b)\u00a0$\\frac{2}{\\sqrt{3}}$<\/p>\n

c)\u00a0$\\frac{4}{\\sqrt{3}}$<\/p>\n

d)\u00a0$\\frac{8}{\\sqrt{3}}$<\/p>\n

Question 9:\u00a0<\/b>$a\\cos60$=$b\\tan60$ and $c\\sec30$=$d\\cot60$,value of ac\/bd is<\/p>\n

a)\u00a0$\\frac{4}{\\sqrt{3}}$<\/p>\n

b)\u00a0$\\frac{2}{\\sqrt{3}}$<\/p>\n

c)\u00a0$\\frac{8}{\\sqrt{3}}$<\/p>\n

d)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n

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Question 10:\u00a0<\/b>$a\\sin30$=$b\\csc60$ and $c\\tan45$=$d\\sec60$,value of ac\/bd is<\/p>\n

a)\u00a0$\\frac{4}{\\sqrt{3}}$<\/p>\n

b)\u00a0$\\frac{2}{\\sqrt{3}}$<\/p>\n

c)\u00a0$\\frac{8}{\\sqrt{3}}$<\/p>\n

d)\u00a0$\\frac{16}{\\sqrt{3}}$<\/p>\n

Question 11:\u00a0<\/b>Find the value of $tan 48^\\circ – tan 3^\\circ – tan 48^\\circ.tan 3^\\circ$<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a0-1<\/p>\n

d)\u00a02<\/p>\n

Question 12:\u00a0<\/b>Find the value of $tan 1^\\circ . tan 2^\\circ . tan 3^\\circ . . . . tan 89^\\circ$<\/p>\n

a)\u00a00<\/p>\n

b)\u00a0-1<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0$\\sqrt{3}$<\/p>\n

Question 13:\u00a0<\/b>A person observes the top of a tower at an angle of 45\u00b0. He walks for 20 m towards the tower and the angle of elevation changes to 60\u00b0. Find the height of the tower?<\/p>\n

a)\u00a0$ 30 + 10\\sqrt{3}$m<\/p>\n

b)\u00a0$ 20 + 10\\sqrt{5}$m<\/p>\n

c)\u00a0$ 30 $m<\/p>\n

d)\u00a0$ 30 + 15\\sqrt{3}$<\/p>\n

Question 14:\u00a0<\/b>If cos x = $\\dfrac{12}{13}$, then find the value of tan x.<\/p>\n

a)\u00a0$\\dfrac{5}{13}$<\/p>\n

b)\u00a0$\\dfrac{5}{12}$<\/p>\n

c)\u00a0$\\dfrac{13}{12}$<\/p>\n

d)\u00a0$\\dfrac{9}{13}$<\/p>\n

Question 15:\u00a0<\/b>$1\/sinA – SinA\/Cos^2A$ = ?<\/p>\n

a)\u00a0$2Cot2A.secA$<\/p>\n

b)\u00a0$2Cos2A.cosecA$<\/p>\n

c)\u00a0$2Cos2A.secA$<\/p>\n

d)\u00a0$2Cot2A.cosecA$<\/p>\n

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Question 16:\u00a0<\/b>If sin x = $\\dfrac{3}{4} ( 0 < \\theta < 90^\\circ )$, find the value of tan x.<\/p>\n

a)\u00a0$\\dfrac{4}{5}$<\/p>\n

b)\u00a0$\\dfrac{3}{4}$<\/p>\n

c)\u00a0$\\dfrac{5}{4}$<\/p>\n

d)\u00a0$\\dfrac{5}{3}$<\/p>\n

Question 17:\u00a0<\/b>FInd sec\u03b8-tan\u03b8, if sec\u03b8+tan\u03b8 = 1\/3 ?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

Question 18:\u00a0<\/b>Find cosec\u03b8, if sin2\u03b8=$\\frac{\\sqrt3}{2}$?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

Question 19:\u00a0<\/b>What is the value of sec\u03b8+cos\u03b8, if sin\u03b8 = 12\/13 ?It is known that 0 < \u03b8 < 90<\/p>\n

a)\u00a0194\/65<\/p>\n

b)\u00a0124\/25<\/p>\n

c)\u00a0169\/25<\/p>\n

d)\u00a0145\/25<\/p>\n

Question 20:\u00a0<\/b>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\u00b0 and 30\u00b0 respectively ? (take tan 35\u00b0=0.7)<\/p>\n

a)\u00a010 meters<\/p>\n

b)\u00a012 meters<\/p>\n

c)\u00a013 meters<\/p>\n

d)\u00a015 meters<\/p>\n

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Question 21:\u00a0<\/b>$sin^{-1}x + cos^{-1}x = ? (-1\\leq x \\leq 1)$<\/p>\n

a)\u00a0$\\pi\/2$<\/p>\n

b)\u00a0$\\pi$<\/p>\n

c)\u00a0$\\pi\/3$<\/p>\n

d)\u00a0$2\\pi$<\/p>\n

Question 22:\u00a0<\/b>If tan(\u03b8\/2) = 5\/12, then find sin\u03b8 = ?<\/p>\n

a)\u00a05\/13<\/p>\n

b)\u00a0119\/120<\/p>\n

c)\u00a0120\/169<\/p>\n

d)\u00a012\/13<\/p>\n

Question 23:\u00a0<\/b>It is known that 0 < \u03b8 < 90 and it is given that cot \u03b8 = 5\/12. Find the value of cosec\u03b8<\/p>\n

a)\u00a05\/13<\/p>\n

b)\u00a013\/12<\/p>\n

c)\u00a012\/5<\/p>\n

d)\u00a012\/13<\/p>\n

Question 24:\u00a0<\/b>If sec \u03b8 + tan \u03b8 = -2, then find sec \u03b8 – tan \u03b8 = ?<\/p>\n

a)\u00a0-1\/2<\/p>\n

b)\u00a02<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a0-2<\/p>\n

Question 25:\u00a0<\/b>Find tan \u03b8, if sec\u03b8 = 13\/12 ?
\nIt is known that 0 < \u03b8 < 90<\/p>\n

a)\u00a012\/5<\/p>\n

b)\u00a05\/12<\/p>\n

c)\u00a012\/13<\/p>\n

d)\u00a05\/13<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

cos 1^\\circ.cos 2^\\circ.cos 3^\\circ.cos 4^\\circ\u2026…cos 179^\\circ
\ncos 1^\\circ.cos 2^\\circ.cos 3^\\circ.cos 4^\\circ\u2026….cos 90^\\circ…..cos 179^\\circ
\nSince cos 90^\\circ = 0, the given series multiplication equals to zero.<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Given $sin x = \\dfrac{9}{41}$
\n\u21d2 $sin^2 x = \\dfrac{81}{1681}$<\/p>\n

We know that $cos^2 x = 1-sin^2 x$
\nThen, $cos^2 x = 1-\\dfrac{81}{1681}$
\n\u21d2 $cos^2 x = \\dfrac{1600}{1681}$<\/p>\n

\u21d2 $cos x = \\dfrac{40}{41}$
\nThen, $sec x = \\dfrac{41}{40}$<\/p>\n

$tan x = \\dfrac{sin x}{cos x}$<\/p>\n

$= \\drac{9}{40}$<\/p>\n

$sec x – tan x = \\dfrac{41}{40} – \\drac{9}{40} = \\dfrac{32}{40} = \\dfrac{4}{5}$<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given $sin x = \\dfrac{7}{25}$
\n\u21d2 $sin^2 x = \\dfrac{49}{625}$
\nWe know that $cos^2 x = 1-sin^2 x$
\n$cos^2 x = 1-\\dfrac{49}{625}$
\n$= \\dfrac{576}{625} = \\dfrac{24}{25}$<\/p>\n

$tan x = \\dfrac{sin x}{cos x}$<\/p>\n

$= \\dfrac{(\\dfrac{7}{25})}{\\dfrac{24}{25}}$<\/p>\n

$= \\dfrac{7}{24}$<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

The below diagram represents the given situation.<\/p>\n

Here, Height of the tower = y metres
\nFrom the figure,
\n$tan 45^\\circ = \\dfrac{y}{x+30}$
\n\u21d2 $1 = \\dfrac{y}{x+30}$
\n\u21d2 $y = x+30$
\n\u21d2 $x = y-30$
\n$tan 60^\\circ = \\dfrac{y}{x}$
\n$\\sqrt{3} = \\dfrac{y}{x}$
\nSubstituting x = y-30 in above equation
\n$\\dfrac{y}{y-30} = \\sqrt{3}$
\n$(y-30)\\sqrt{3} = y$
\n$\\sqrt{3}y-y = 30\\sqrt{3}$
\n\u21d2 $y(\\sqrt{3}-1) = 30\\sqrt{3}$
\n\u21d2 $y = \\dfrac{30\\sqrt{3}}{\\sqrt{3}-1}$
\nRationalising above equation
\n$y = \\dfrac{30\\sqrt{3}}{\\sqrt{3}-1} \\times \\dfrac{\\sqrt{3}+1}{\\sqrt{3}+1}$<\/p>\n

$y = \\dfrac{30\\sqrt{3}(\\sqrt{3}+1}{2} = \\dfrac{90+30\\sqrt{3}}{2} = 45+15\\sqrt{3}<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

We know that $\\sin\\theta$=$\\cos(90-\\theta)$
\nSo for $\\sin20$ we have $\\cos70$
\n$\\sin40$ we have $\\cos50$
\n$\\sin60$ we have $\\cos30$
\nTherefore require value is 1<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$(\\sin^{2}x+\\cos^{2}x)+\\sin^{2}x$=1+$\\sin^{2}x$
\nMinimum value of $\\sin x$ is when $x$=0 i.e $\\sin 0$=0
\nTherefore 1+0=1<\/p>\n

7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

12$(\\sin^{2}x+\\cos^{2}x)+\\cos^{2}x$=12(1)+$\\cos^{2}x$
\nMaximum value of $\\cos x$ is when $x$=0 i.e $\\cos x$=1
\nTherefore 12+1=13<\/p>\n

8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

we have a=($b\\cos60*\\cos60$)
\na=b($\\frac{1}{4}$)
\n(a\/b)=$\\frac{1}{4}$
\n$(c\/\\sin30$)=$\\frac{d}{\\sin60}$
\n(c\/d)=$\\frac{4}{\\sqrt{3}}$<\/p>\n

(ac\/bd)=$\\frac{1}{\\sqrt{3}}$<\/p>\n

9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

we have (a\/2)=($b\\tan60$)<\/p>\n

(a\/2)=$\\frac{b}{\\sqrt{3}}$<\/p>\n

(a\/b)=$\\frac{2}{\\sqrt{3}}$<\/p>\n

$(c\/\\cos30$)=$\\frac{d}{\\sqrt{3}}$<\/p>\n

$\\frac{2c}{\\sqrt{3}}$=$\\frac{d}{\\sqrt{3}}$<\/p>\n

(c\/d)=(1\/2)
\n(ac\/bd)=$\\frac{1}{\\sqrt{3}}$<\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

we have (a\/2)=(b\/$\\sin60$)
\n(a\/2)=$\\frac{2b}{\\sqrt{3}}$
\n(a\/b)=$\\frac{4}{\\sqrt{3}}$
\nc=$\\frac{d}{\\cos60}$
\nc=2d
\n(c\/d)=2
\n(ac\/bd)=$\\frac{8}{\\sqrt{3}}$<\/p>\n

\"RRB<\/a><\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$48-3 = 45$
\nApplying tan on both sides
\n$tan (48-3)^\\circ = tan 45^\\circ$ \u2192 (1)
\nWe know that $tan (A-B)^\\circ = \\dfrac{tan A^\\circ – tan B^\\circ}{1+tan A^\\circ.tan B^\\circ}$
\nThen, (1) becomes
\n$\\dfrac{tan 48^\\circ – tan 3^\\circ}{1+tan 48^\\circ.tan 3^\\circ} = 1$ (Since, $tan 45^\\circ = 1$)<\/p>\n

$tan 48^\\circ – tan 3^\\circ = 1+tan 48^\\circ.tan 3^\\circ$
\n\u21d2 $tan 48^\\circ – tan 3^\\circ – tan 48^\\circ.tan 3^\\circ = 1$<\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

The given equation can be written as
\n$tan 1^\\circ . tan 89^\\circ . tan 2^\\circ . tan 88^\\circ . . . . tan 45^\\circ$
\n$= tan 1^\\circ . tan (90-1)^\\circ . tan 2^\\circ . tan (90-2)^\\circ . . . . tan 45^\\circ$
\nWe know that $tan (90-x)^\\circ = cot x^\\circ$
\nThen, the above equation becomes
\n$tan 1^\\circ . cot 1^\\circ . tan 2^\\circ . cot 2^\\circ . . . . tan 45^\\circ$
\nSince tan x^\\circ. cot x^\\circ = 1, the above equation equals to 1.
\nHence, $tan 1^\\circ . tan 2^\\circ . tan 3^\\circ . . . . tan 89^\\circ = 1$<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

The given situation can be represented through the following figure.<\/p>\n

<\/figure>\n

We know that
\n$Tan 45^\\circ =$ h\/BC
\n=> 1 = h\/BC
\n=> BC = h
\nNow after walking for 20m, the angle changed to 60\u00b0.
\nSo we have
\nTan 60 = h\/BD
\n=> $ \\sqrt{3} = \\frac{h}{h – 20}$
\n=> $20\\sqrt{3} = h(\\sqrt{3} – 1)$
\n=> h = $\\frac{20\\sqrt{3}}{\\sqrt{3} – 1}$
\n=> h = $ \\frac{60 + 20\\sqrt{3}}{2}$
\n=> h = $ 30 + 10\\sqrt{3}$<\/p>\n

14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given cos x = $\\dfrac{12}{13}$<\/p>\n

Squaring on both sides
\n$cos^{2}x = \\dfrac{144}{169}$<\/p>\n

We know that $sin^2 x = 1-cos^2 x$
\n\u21d2 $sin^2 x = 1-$\\dfrac{144}{169}$<\/p>\n

\u21d2 $sin^2 x = $\\dfrac{25}{169}$<\/p>\n

\u21d2 $sin x = $\\dfrac{5}{13}$<\/p>\n

We know that tan x = $\\dfrac{sin x}{cos x}$<\/p>\n

tan x $= \\dfrac(\\dfrac{5}{13})}{(\\dfrac{12}{13})} = \\dfrac{5}{12}$<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$1\/sinA – SinA\/Cos^2A$
\n= $(cos^2A-sin^2A)\/(sinA.cos^2A)$
\n= $2(Cos2A)\/(sin2A.CosA)$
\n= $2Cot2A.secA$
\nSo the answer is option A.<\/p>\n

16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

We know that cos x = $\\sqrt{1-sin^{2}x}$
\n$= \\sqrt{1-\\dfrac{9}{25}}$<\/p>\n

$= \\sqrt{\\dfrac{16}{25}} = \\dfrac{4}{5}$<\/p>\n

tan x = \\dfrac{sin x}{cos x} = \\dfrac{(\\dfrac{3}{5})}{(\\dfrac{4}{5})} = \\dfrac{3}{4}$<\/p>\n

17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

W.K.T $sec^2\\theta – tan^2\\theta = 1$<\/p>\n

$(sec\\theta-tan\\theta)(sec\\theta+tan\\theta) = 1$<\/p>\n

$(sec\\theta-tan\\theta)(1\/3) = 1$<\/p>\n

$sec\\theta-tan\\theta = 3$<\/p>\n

So the answer is option D.<\/p>\n

18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

sin2\u03b8=$\\frac{\\sqrt3}{2}$ then 2\u03b8 = 60\u00b0 ==> \u03b8 = 30\u00b0<\/p>\n

cosec\u03b8 = cosec30\u00b0 = 2<\/p>\n

So the answer is option C.<\/p>\n

19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

If sin\u03b8 = 12\/13 cos\u03b8 = 5\/13 and sec\u03b8 = 13\/5<\/p>\n

sec\u03b8+cos\u03b8 = 13\/5 + 5\/13 = 194\/65<\/p>\n

So the answer is option A.<\/p>\n

20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

<\/figure>\n

Tan35\u00b0 = 49\/y<\/p>\n

0.7 = 49\/y<\/p>\n

y = 70<\/p>\n

Tan30\u00b0 = 49\/(x+y)<\/p>\n

1\/\u221a3 = 49\/(x+70)<\/p>\n

x+70 = 49\u221a3<\/p>\n

x+70 = 84.87<\/p>\n

x = 84.87-70 = 14.87 $\\sim$ 15 meters<\/p>\n

So the answer is option D.<\/p>\n

21)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

For any value of x, $sin^{-1}x + cos^{-1}x = \\pi\/2$<\/p>\n

Lets take x = 1\/2<\/p>\n

$sin^{-1}(1\/2) + cos^{-1}(1\/2) = 30 + 60 = 90$\u00b0= $\\pi\/2$<\/p>\n

So the answer is option A.<\/p>\n

22)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$Sin\u03b8$ = $\\frac{2tan(\u03b8\/2)}{1+tan^2(\u03b8\/2)}$ = $\\frac{2(5\/12)}{1+(5\/12)^2}$ = $120\/169$<\/p>\n

So the answer is option C.<\/p>\n

23)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$\\text{cot} \\theta = \\frac{5}{12}$<\/p>\n

$\\text{cosec}^2 \\theta$ = $1+\\text{cot}^2 \\theta = 1+{(\\frac{5}{12})}^2 = \\frac{169}{144}$<\/p>\n

$\\text{cosec}\\theta = \\frac{13}{12}$<\/p>\n

So the answer is option B.<\/p>\n

24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$sec^2 \u03b8 – tan^2 \u03b8 = 1$<\/p>\n

$(sec \u03b8 – tan \u03b8)(sec \u03b8 + tan \u03b8) = 1$<\/p>\n

$(sec \u03b8 – tan \u03b8) (-2) = 1$<\/p>\n

$(sec \u03b8 – tan \u03b8) = -1\/2$<\/p>\n

So the answer is option A.<\/p>\n

\"RRB<\/a><\/p>\n

25)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$sec^2\u03b8-tan^2\u03b8 = 1$<\/p>\n

$(13\/12)^2-tan^2\u03b8 = 1$<\/p>\n

$169\/144-tan^2\u03b8 = 1$<\/p>\n

$tan^2\u03b8 = 25\/144$<\/p>\n

$tan\u03b8 = 5\/12$<\/p>\n

So the answer is option B.<\/p>\n

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We hope this Trigonometry Questions for RRB JE Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

RRB JE Trigonometry Questions PDF Download RRB JE Trigonometry\u00a0 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:\u00a0Find the value of cos 1^\\circ.cos 2^\\circ.cos 3^\\circ.cos 4^\\circ\u2026…cos 179^\\circ a)\u00a0-1 b)\u00a00 c)\u00a0$\\sqrt{2}$ d)\u00a01 Question 2:\u00a0If sin x $= \\dfrac{9}{41}$ […]<\/p>\n","protected":false},"author":41,"featured_media":26097,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1605],"tags":[489,491,1635,1647],"class_list":{"0":"post-26094","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-je","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-je","14":"tag-rrb-mocks"},"better_featured_image":{"id":26097,"alt_text":"RRB JE Trigonometry 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