{"id":27423,"date":"2019-04-12T18:41:45","date_gmt":"2019-04-12T13:11:45","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=27423"},"modified":"2019-04-15T10:35:49","modified_gmt":"2019-04-15T05:05:49","slug":"rrb-group-d-trigonometry-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/rrb-group-d-trigonometry-questions-pdf\/","title":{"rendered":"RRB Group-D Trigonometry Questions PDF"},"content":{"rendered":"

RRB Group-D Trigonometry Questions PDF<\/span><\/h1>\n

Download Top 20 RRB Group-D Trigonometry Questions and Answers PDF. RRB Group-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.<\/p>\n

Download RRB Group-D Trigonometry Questions PDF<\/a><\/p>\n\n

Download RRB Group-D Previous Papers PDF<\/a><\/p>\n

Take a RRB Group-D free mock test<\/a><\/p>\n

Question 1:\u00a0<\/b>$5tan\\theta = 4$, then the value of\u00a0$(\\frac{5sin\\theta – 3cos\\theta}{5sin\\theta + 3cos\\theta})$ is<\/p>\n

a)\u00a0$\\frac{1}{7}$<\/p>\n

b)\u00a0$\\frac{2}{7}$<\/p>\n

c)\u00a0$\\frac{5}{7}$<\/p>\n

d)\u00a0$\\frac{2}{5}$<\/p>\n

Question 2:\u00a0<\/b>The least value of $(4sec^2\\theta + 9cosec^2\\theta)$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a019<\/p>\n

c)\u00a025<\/p>\n

d)\u00a07<\/p>\n

Question 3:\u00a0<\/b>If $x=cosec\\theta-sin\\theta$ and $y=sec\\theta-cos\\theta$, then the value of $x2<\/sup>y2<\/sup>(x2<\/sup> + y2<\/sup> + 3)$<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

Question 4:\u00a0<\/b>If $ 0 \\leq \\theta \\leq \\frac{\\pi}{2}$, $2ycos\\theta=sin\\theta$ and $\\frac{x}{2cosec\\theta}=y$, then the value of $x^2-4y^2$\u00a0is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a04<\/p>\n

Question 5:\u00a0<\/b>If $sin\\theta + \\sin^2\\theta = 1$, then the value of cos12<\/sup>$\\theta$ + 3cos10<\/sup>$\\theta$ + cos6<\/sup>$\\theta$ + \u00a03cos8<\/sup>$\\theta$\u00a0 – 1\u00a0is<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

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

d)\u00a02<\/p>\n

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Question 6:\u00a0<\/b>The value of $\\frac{1}{cosec\\theta – cot\\theta} – \\frac{1}{sin\\theta}$<\/p>\n

a)\u00a0$cot\\theta$<\/p>\n

b)\u00a0$cosec\\theta$<\/p>\n

c)\u00a0$tan\\theta$<\/p>\n

d)\u00a0$1$<\/p>\n

Question 7:\u00a0<\/b>If\u00a0$\\cos\\theta + \\sin\\theta = \\sqrt{2}\\cos\\theta$, then $\\cos\\theta – \\sin\\theta$ is<\/p>\n

a)\u00a0-$\\sqrt{2}\\cos\\theta$<\/p>\n

b)\u00a0-$\\sqrt{2}\\sin\\theta$<\/p>\n

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

d)\u00a0$\\sqrt{2}\\tan\\theta$<\/p>\n

Question 8:\u00a0<\/b>If $cos^4\\theta-sin^4\\theta=\\frac{2}{3}$, then the value of $1-2sin^2\\theta$ is,<\/p>\n

a)\u00a00<\/p>\n

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

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

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

Question 9:\u00a0<\/b>If $tan\\theta$ = 3\/4 and $\\theta$ is acute, then $cosec\\theta$ is equal to<\/p>\n

a)\u00a0$\\frac{5}{3}$<\/p>\n

b)\u00a0$2$<\/p>\n

c)\u00a0$\\frac{1}{2}$<\/p>\n

d)\u00a0$4$<\/p>\n

Question 10:\u00a0<\/b>The value of $\\frac{1}{1 + tan^2\\theta}$ + $\\frac{1}{1 + cot^2\\theta}$ is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a0$\\frac{1}{2}$<\/p>\n

d)\u00a0$\\frac{1}{4}$<\/p>\n

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Question 11:\u00a0<\/b>Maximum value of $(2sin\\theta+3 cos\\theta)$ is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a0$\\sqrt{13}$<\/p>\n

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

d)\u00a01<\/p>\n

Question 12:\u00a0<\/b>If $Cos \\Theta + Sec\\Theta = 2$, find $Cos^{2017} \\Theta + Sec^{2017}\\Theta $<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a01018<\/p>\n

d)\u00a02017<\/p>\n

Question 13:\u00a0<\/b>If $\\sin x+\\frac{1}{\\sin x}=2$ then $\\cos ^5 x+\\cot ^{5} x=?$<\/p>\n

a)\u00a00<\/p>\n

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

c)\u00a01<\/p>\n

d)\u00a02<\/p>\n

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Question 14:\u00a0<\/b>If $cos^2x + (1 + sinx)^2 = 3$, then find the value of cosec x + sin(60+x), where 0\u00ba < x < 90\u00ba?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a02<\/p>\n

c)\u00a0$2\\frac{1}{2}$<\/p>\n

d)\u00a0$3\\frac{1}{2}$<\/p>\n

Question 15:\u00a0<\/b>If $x$ and $y$ are acute angles such that their sum is less than $90^{\\circ}$ . If $\\cos(2x-20^{\\circ})=\\sin(2y+20^{\\circ})$ , then the value of $\\tan(x+y)$ is<\/p>\n

a)\u00a0$0$<\/p>\n

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

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

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

Question 16:\u00a0<\/b>If x is an acute angle such that
\nTan (4x – 50\u00b0) = Cot ( 50\u00b0 – x), then the value of x would be?<\/p>\n

a)\u00a060<\/p>\n

b)\u00a045<\/p>\n

c)\u00a050<\/p>\n

d)\u00a030<\/p>\n

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Question 17:\u00a0<\/b>If cosec 9x= sec 9x (0 <x<10), what is the value of x?<\/p>\n

a)\u00a09\u00b0<\/p>\n

b)\u00a03\u00b0<\/p>\n

c)\u00a05\u00b0<\/p>\n

d)\u00a06\u00b0<\/p>\n

Question 18:\u00a0<\/b>If $\\sin x+1=\\frac{2}{\\sin x} $ then $\\cos ^3 x+ \\mathrm{cosec} ^{-3} x=?$<\/p>\n

a)\u00a00<\/p>\n

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

c)\u00a01<\/p>\n

d)\u00a02<\/p>\n

Question 19:\u00a0<\/b>If $\\frac{sin2x}{sinx} + cos2x + sin^2x = 0$, find the value of x, if $0 \\leq x \\leq 180$.<\/p>\n

a)\u00a045\u00ba<\/p>\n

b)\u00a090\u00ba<\/p>\n

c)\u00a0180\u00ba<\/p>\n

d)\u00a0More than one value is possible<\/p>\n

Question 20:\u00a0<\/b>If $\\frac{sin^2 \\Theta}{4} = \\frac{cos^2 \\Theta}{9}$. Find the value of $tan^2 \\Theta – cot^2 \\Theta$.<\/p>\n

a)\u00a01\/6<\/p>\n

b)\u00a013\/27<\/p>\n

c)\u00a01\/18<\/p>\n

d)\u00a0None of these<\/p>\n

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

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

Taking $cos\\theta$ outside in numerator and in denominator and making $tan\\theta$
\nhence eq will be\u00a0\u00a0$(\\frac{5tan\\theta – 3}{5tan\\theta + 3})$
\nAs it is given that $5tan\\theta$ = 4
\nafter putting values and solving we will get the equation reduced to 1\/7.<\/p>\n

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

$4sec^2\\theta+9cosec^2\\theta$
\nor $4+4tan^2\\theta+9+9cot^2\\theta$
\nor $13+4tan^2\\theta+9cot^2\\theta$
\nor $ 13+4tan^2\\theta+\\frac{9}{tan^2\\theta} $
\nor $ \u00a013-12+(2tan\\theta+\\frac{3}{tan\\theta})^2 $ \u00a0 \u00a0(eq. (1) )
\nor now above expression to be minimum, equation $(2tan\\theta+\\frac{3}{tan\\theta})^2$ should be minimum.
\nSo applying $A.M.\\geq G.M. $
\n$\\frac{(2tan\\theta +\\frac{3}{tan\\theta})}{2} \\geq \\sqrt{6}$
\nor\u00a0${(2tan\\theta+\\frac{3}{tan\\theta})}=2\\sqrt{6}$ ( for value to be minimum)
\nAfter putting above value in eq.(1) , we will get least value of expression as 25.<\/p>\n

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

$x=cosec\\theta – sin\\theta=\\frac{cos^2\\theta}{sin\\theta}=cot\\theta cos\\theta$
\nSimilarly $y=tan\\theta sin\\theta$
\n$xy=sin\\theta cos\\theta$
\n$x^2+y^2+3=(sec^2\\theta +cosec^2\\theta )$
\nNow putting above values in given equation, and after solving it will be reduced to 1<\/p>\n

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

$2y=tan\\theta$
\n$x=2ycosec\\theta$
\nHence value of $x^2 – 4y^2 $ = $4y^2(cosec^2\\theta – 1)$
\nor $tan^2\\theta cot^2\\theta$ = 1<\/p>\n

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

Given equation can be written as $(cos^4\\theta + cos^2\\theta)^3 -1$
\nas $sin\\theta + sin^2\\theta = 1$
\nor $sin\\theta = cos^2\\theta$
\nputting above value in given equation it will be
\n$(sin^2\\theta + sin\\theta)^3 -1 = 0$<\/p>\n

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

$\\frac{sin\\theta}{1-cos\\theta} – \\frac{1}{sin\\theta}$<\/p>\n

or $\\frac{cos\\theta – cos^2\\theta}{(1-cos\\theta)sin\\theta}$ = $cot\\theta$<\/p>\n

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

$\\sin^2 \\theta + \\cos^2 \\theta = 1$
\nSo, $\\sin^2 \\theta + \\cos^2 \\theta + 2\\sin\\theta * \\cos \\theta = 2 \\cos^2\\theta$
\nHence, $\\cos^2 \\theta – \\sin^2 \\theta = 2 \\sin\\theta*\\cos\\theta$
\nSo, $\\cos\\theta – \\sin\\theta = \\sqrt{2}\\sin\\theta$<\/p>\n

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

$cos^4\\theta-sin^4\\theta=(cos^2\\theta-sin^2\\theta)(cos^2\\theta+sin^2\\theta)=cos^2\\theta-sin^2\\theta=\\frac{2}{3}$
\n$cos^2\\theta-sin^2\\theta =1-2sin^2\\theta=\\frac{2}{3}$<\/p>\n

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

$\\frac{sin \\theta}{cos \\theta} = \\frac{3}{4}$
\nSo, $\\frac{sin^2\\theta}{cos^2\\theta}=\\frac{9}{16}$
\nHence, $sin^2 \\theta = \\frac{9}{9+16}=\\frac{9}{25}$
\nSo, $cosec \\theta = \\frac{5}{3}$<\/p>\n

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

$1 + \\tan ^2 \\theta = \\sec ^2 \\theta$
\n$1 + \\cot ^2 \\theta = \\csc ^2 \\theta$
\nSo, the given fraction becomes,<\/p>\n

$\\frac{1}{\\sec ^2 \\theta} + \\frac{1}{\\csc ^2 \\theta} = \\sin^2\\theta + \\cos^2 \\theta = 1$<\/p>\n

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

$\\because$ Maximum Value of $a\\sin{\\theta}+b\\cos{\\theta}=\\sqrt{a^{2}+b^{2}}$
\n$\\therefore$ Maximum Value of $2\\sin{\\theta}+3\\cos{\\theta}=\\sqrt{2^{2}+3^{2}}$
\n$=\\sqrt{13}$
\nHence, Correct option is B.<\/p>\n\n

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

$Cos \\Theta + Sec\\Theta = 2$
\n$(Cos \\Theta + Sec\\Theta )^2 = 2^2$
\n==> $Cos^2 \\Theta + Sec^2\\Theta + 2 Cos \\Theta * Sec \\Theta = 4$ (Note :Cos \\Theta * Sec \\Theta = 1)
\n==> $Cos^2 \\Theta + Sec^2\\Theta +2 = 4$
\n==> $Cos^2 \\Theta + Sec^2\\Theta $ = 2
\n$(Cos \\Theta + Sec\\Theta )^3 = 2^3$
\n==> $(Cos^3 \\Theta + Sec^3\\Theta + 3*Cos \\Theta * Sec \\Theta(Cos \\Theta + Sec\\Theta) = 2^3$
\n==> $(Cos^3 \\Theta + Sec^3\\Theta$ = 2
\nTherefore, irrespective of the power, the expression always equals to 2.<\/p>\n

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

$\\sin x+\\frac{1}{\\sin x}=2$
\n=>$\\sin^2 x+1=2\\sin x$
\n=>$(\\sin x-1)^2=0$
\n=>$\\sin x = 1$
\n=>$\\cos x = 0$
\nThe given expression would be,
\n$\\cos ^5 x+\\cot ^{5} x$
\n$\\cos ^5 x+\\frac{\\cos^5 x}{\\sin^5 x}$
\n$=0$<\/p>\n

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

$cos^2x + (1 + sinx)^2 = 3$
\n=> $cos^2x + 1 + sin^2x + 2sinx = 3$ (Since $ cos^2x + sin^2x = 1$)
\n=> $2 sin x = 1$ or sin x = \u00bd which means x=30 degrees [Solution in the first quadrant]
\n=> cosec x = 1\/sin x = 2 and sin (60+x) = sin 90 = 1
\n=> The value of the expression is 2+1 = 3<\/p>\n

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

Given that , $x+y<90^{\\circ} \\dots (1)$
\n$\\implies\\cos(2x-20^{\\circ})=\\cos(90^{\\circ}-(2y+20^{\\circ}))$
\nUsing the given property in $(1)$
\n$\\implies 2x-20^{\\circ}=70^{\\circ}-2y$
\n$\\implies 2x+2y=90^{\\circ}$
\n$\\implies x+y=45^{\\circ}$
\n$\\implies \\tan(x+y)=1 $
\nHence , the correct option is B<\/p>\n

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

We know that Tan ( 90 – x) = Cot x and Cot (90 – x) = Tanx
\nSo 4x – 50 = 90 – y
\nThen 50 – x will be y
\n=> y = 50 – x
\nx + y = 50
\nMoreover,
\n90 – y = 4x – 50
\n=> 4x + y = 140
\nSubtracting the two equations, we get
\n3x = 90
\n=> x = 30
\nHence the correct answer is option D.<\/p>\n

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

If cosec 9x= sec 9x,
\nThen $\\sin 9x=\\cos 9x$
\nIn the first quadrant cos and sin are equal when $9x =45$\u00b0
\nHence, x = 5\u00b0<\/p>\n

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

$\\sin x+1=\\frac{2}{\\sin x} $
\n=>$\\sin^2 x+\\sin x=2$
\n=>$(\\sin x-1)(\\sin x+2)=0$
\nAs $\\sin x$ can only take values between -1 and 1,
\n=>$\\sin x = 1$
\n=>$\\cos x = 0$
\nThe given expression would be,
\n$\\cos ^3 x+ \\mathrm{cosec}^{-3} x$
\n$\\cos ^3 x + \\sin^3 x$
\n$=1$<\/p>\n

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

$\\frac{sin2x}{sinx} + cos2x + sin^2x = 0$
\nWe know that $sin2x = 2 sinx cosx$ and $cos2x=cos^2x-sin^2x$
\n=> $2cosx + cos^2x = 0$
\n=> cos x =0 or cos x =2
\nNow, cos x =2 is not possible.
\n=> cos x = 0
\nIn the range, $0 \\leq x \\leq 180$, cos x =0 at only 90\u00ba
\nThus, B is the correct answer.<\/p>\n

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

$\\frac{sin^2 \\Theta}{4} = \\frac{cos^2 \\Theta}{9}$
\n==> $\\frac{sin^2 \\Theta}{cos^2 \\Theta}$ = $\\frac{4}{9}$
\n==> $tan^2 \\Theta = \\frac{4}{9}$
\n==> $cot^2 \\Theta = \\frac{9}{4}$
\n==> $tan^2 \\Theta – cot^2 \\Theta$ = 4\/9 – 9\/4 = -65\/36
\nSince there is no such option, the correct option to choose is D.<\/p>\n\n

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

RRB Group-D Trigonometry Questions PDF Download Top 20 RRB Group-D Trigonometry Questions and Answers PDF. RRB Group-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam. Download RRB Group-D Previous Papers PDF Take a RRB Group-D free mock test Question 1:\u00a0$5tan\\theta = 4$, then the value of\u00a0$(\\frac{5sin\\theta […]<\/p>\n","protected":false},"author":41,"featured_media":27426,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[31,1679],"tags":[489,490,491,1647],"class_list":{"0":"post-27423","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-railways","8":"category-rrb-group-d","9":"tag-railway-exam","10":"tag-railway-group-d","11":"tag-rrb","12":"tag-rrb-mocks"},"better_featured_image":{"id":27426,"alt_text":"RRB Group-D Trigonometry Questions PDF","caption":"RRB Group-D Trigonometry Questions PDF","description":"RRB Group-D Trigonometry Questions 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