{"id":28181,"date":"2019-04-29T15:44:58","date_gmt":"2019-04-29T10:14:58","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=28181"},"modified":"2019-04-29T15:44:58","modified_gmt":"2019-04-29T10:14:58","slug":"trigonometry-questions-for-rrb-ntpc-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/trigonometry-questions-for-rrb-ntpc-pdf\/","title":{"rendered":"Trigonometry Questions for RRB NTPC PDF"},"content":{"rendered":"

Trigonometry Questions for RRB NTPC PDF<\/strong><\/span><\/h1>\n

Download RRB NTPC Trigonometry Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

Download Trigonometry Questions for RRB NTPC PDF<\/a><\/p>\n\n

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Question 1:\u00a0<\/b>$\\frac{2sin\u03b8}{1+cos\u03b8}$ = ?<\/p>\n

a)\u00a02cot\u03b8\/2<\/p>\n

b)\u00a0cot\u03b8\/2<\/p>\n

c)\u00a0tan\u03b8\/2<\/p>\n

d)\u00a02tan\u03b8\/2<\/p>\n

Question 2:\u00a0<\/b>If $cos\u03b8+\\frac{1}{cos\u03b8}=\\frac{5}{2}$ , then find $cos^{2}\u03b8+\\frac{1}{cos^{2}\u03b8}$ = ?<\/p>\n

a)\u00a025\/4<\/p>\n

b)\u00a04\/25<\/p>\n

c)\u00a04\/17<\/p>\n

d)\u00a017\/4<\/p>\n

Question 3:\u00a0<\/b>$\\frac{sinA+sinB}{cosA-cosB}$ = ?<\/p>\n

a)\u00a0-cot($\\frac{A-B}{2}$)<\/p>\n

b)\u00a0-tan($\\frac{A-B}{2}$)<\/p>\n

c)\u00a0cot($\\frac{A+B}{2}$)<\/p>\n

d)\u00a0tan($\\frac{A+B}{2}$)<\/p>\n

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Question 4:\u00a0<\/b>It is known that 0 <\u00a0\u03b8 < 90 and it is also given that sin \u03b8 = 3\/5. Find the value of tan (\u03b8\/2)<\/p>\n

a)\u00a01\/3<\/p>\n

b)\u00a03<\/p>\n

c)\u00a02\/3<\/p>\n

d)\u00a03\/2<\/p>\n

Question 5:\u00a0<\/b>FInd cot\u03b8+tan\u03b8 if sin\u03b8 = 5\/13<\/p>\n

a)\u00a013\/60<\/p>\n

b)\u00a060\/13<\/p>\n

c)\u00a0169\/60<\/p>\n

d)\u00a060\/169<\/p>\n

Question 6:\u00a0<\/b>What is the maximum value of $\\Large\\frac{24}{25}$ sin \u03b8 + $\\Large\\frac{7}{25}$ cos \u03b8 + 5?<\/p>\n

a)\u00a04<\/p>\n

b)\u00a06<\/p>\n

c)\u00a05<\/p>\n

d)\u00a03<\/p>\n

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Question 7:\u00a0<\/b>What is the minimum value of 3 sin \u03b8 + 4 cos \u03b8 ?<\/p>\n

a)\u00a025<\/p>\n

b)\u00a0-25<\/p>\n

c)\u00a05<\/p>\n

d)\u00a0-5<\/p>\n

Question 8:\u00a0<\/b>If cos \u03b8 = 40\/41, then find the value of cot \u03b8 + tan \u03b8 ( 0 <\u00a0\u03b8 < $90^\\circ$) ?<\/p>\n

a)\u00a0$\\Large\\frac{1680}{360}$<\/p>\n

b)\u00a0$\\Large\\frac{360}{1680}$<\/p>\n

c)\u00a0$\\Large\\frac{1681}{360}$<\/p>\n

d)\u00a0$\\Large\\frac{360}{1681}$<\/p>\n

Download General Science Notes PDF<\/a><\/p>\n\n

Question 9:\u00a0<\/b>If cosec \u03b8 – cot \u03b8 = 3, then find cosec \u03b8 + cot \u03b8 = ?<\/p>\n

a)\u00a0$\\Large\\frac{1}{9}$<\/p>\n

b)\u00a0$\\Large\\frac{1}{27}$<\/p>\n

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

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

Question 10:\u00a0<\/b>$ \\Large\u00a0\\frac{sin 75 + sin 15}{cos 75 – cos 15}$ = ?<\/p>\n

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

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

c)\u00a0$-\\sqrt{7}$<\/p>\n

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

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Question 11:\u00a0<\/b>It is known that 0 < \u03b8 < 90 and it is also given that $\\cos \u03b8 = \\frac{24}{25}$, then find the value of $ \\tan(\\frac{\u03b8}{2})$?<\/p>\n

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

b)\u00a0$\\Large\\frac{1}{5}$<\/p>\n

c)\u00a0$\\Large\\frac{1}{24}$<\/p>\n

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

Question 12:\u00a0<\/b>If $sin x = \\dfrac{20}{29}$, then find the value of $sec x – tan x$<\/p>\n

a)\u00a0$\\dfrac{2}{7}$<\/p>\n

b)\u00a0$\\dfrac{4}{7}$<\/p>\n

c)\u00a0$\\dfrac{3}{7}$<\/p>\n

d)\u00a0$\\dfrac{1}{7}$<\/p>\n

Question 13:\u00a0<\/b>If $sin x = \\dfrac{20}{29}$, then find the value of $cosec x – cot x$<\/p>\n

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

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

c)\u00a0$\\dfrac{7}{20}$<\/p>\n

d)\u00a0$\\dfrac{8}{21}$<\/p>\n

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Question 14:\u00a0<\/b>Find the value of $sin^2 1+sin^2 2+sin^2 3 +…..+sin^2 89$<\/p>\n

a)\u00a0$45.5$<\/p>\n

b)\u00a0$44.5$<\/p>\n

c)\u00a0$46.5$<\/p>\n

d)\u00a0$43.5$<\/p>\n

Question 15:\u00a0<\/b>Find the value of $sec 1^\\circ.sec 2^\\circ.sec 3^\\circ\u2026..sec 179^\\circ$.<\/p>\n

a)\u00a00<\/p>\n

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

c)\u00a01<\/p>\n

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

Download Current Affairs Questions & Answers PDF<\/a><\/p>\n

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

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

$\\frac{2sin\u03b8}{1+cos\u03b8}$<\/p>\n

=\u00a0$\\frac{2.2sin\u03b8\/2.cos\u03b8\/2}{2cos^2\u03b8\/2}$<\/p>\n

=\u00a0$\\frac{2sin\u03b8\/2}{cos\u03b8\/2}$<\/p>\n

= $2tan\u03b8\/2$<\/p>\n

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

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

$cos\u03b8+\\frac{1}{cos\u03b8}=5\/2$<\/p>\n

Squaring on both sides<\/p>\n

$(cos\u03b8+\\frac{1}{cos\u03b8})^2=(5\/2)^2$<\/p>\n

$cos^{2}\u03b8+\\frac{1}{cos^{2}\u03b8}+2 = 25\/4$<\/p>\n

$cos^{2}\u03b8+\\frac{1}{cos^{2}\u03b8} = 25\/4-2 = (25-8)\/4 = 17\/4$<\/p>\n

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

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

$\\frac{sinA+sinB}{cosA-cosB}$<\/p>\n

=$\\frac{2sin(\\frac{A+B}{2})cos(\\frac{A-B}{2})}{-2sin(\\frac{A+B}{2})sin(\\frac{A-B}{2})}$<\/p>\n

=-cot($\\frac{A-B}{2}$)<\/p>\n

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

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

Sin\u03b8 = 3\/5 then cos\u03b8 = 4\/5<\/p>\n

$Tan\u03b8\/2 = \\frac{sin\u03b8}{1+cos\u03b8} = \\frac{3\/5}{1+4\/5} = \\frac{3\/5}{9\/5} = \\frac{3}{9} = 1\/3$<\/p>\n

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

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

<\/figure>\n

From the diagram, tan\u03b8 = 5\/12, cot\u03b8 = 12\/5<\/p>\n

cot\u03b8+tan\u03b8 = 12\/5 + 5\/12 = 169\/60<\/p>\n

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

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

For the expression asin \u03b8 + bcos \u03b8 + c,<\/p>\n

The maximum value = $c+\\sqrt{a^{2}+b^{2}}$ &<\/p>\n

The minimum value = $c-\\sqrt{a^{2}+b^{2}}$<\/p>\n

for $\\Large\\frac{24}{25}$ sin \u03b8 + $\\Large\\frac{7}{25}$ cos \u03b8 + 5 , a = $\\Large\\frac{24}{25}$, b = $\\Large\\frac{7}{25}$, c = 5<\/p>\n

maximum value = $ c + \\sqrt{a^{2}+b^{2}}$
\n$= 5 + \\Large\\sqrt{ \\frac{24}{25}^{2} + \\frac{7}{25}^{2} } $<\/p>\n

$= 5+1 = 6$<\/p>\n

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

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

For the expression asin \u03b8 + bcos \u03b8 + c,<\/p>\n

The maximum value = $c+\\sqrt{a^{2}+b^{2}}$ &<\/p>\n

The minimum value = $c-\\sqrt{a^{2}+b^{2}}$<\/p>\n

for 3 sin \u03b8 + 4 cos \u03b8 , a = 3, b = 4, c = 0<\/p>\n

minimum value = $c-\\sqrt{a^{2}+b^{2}}$
\n$= 0 – \\sqrt{3^{2} + 4^{2}} = -\\sqrt{25} = -5$<\/p>\n

so the answer is option d.<\/p>\n

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

Given,
\ncos \u03b8 = $\\Large\\frac{40}{41}$<\/p>\n

sin \u03b8 = $\\Large\\frac{9}{41}$<\/p>\n

tan \u03b8 = $\\Large\\frac{9}{40}$ [As,\u00a0tan \u03b8 = $\\Large\\frac{sin \u03b8}{cos\u00a0\u03b8}$]<\/p>\n

cot \u03b8 = $\\Large\\frac{40}{9}$<\/p>\n

cot \u03b8 + tan \u03b8 = $\\Large\\frac{9}{40}$ + $\\Large\\frac{40}{9}$<\/p>\n

= $\\Large\\frac{1681}{360}$<\/p>\n

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

We know that $ cosec^2 \u03b8 – cot^2 \u03b8 = 1$
\n$ (cosec \u03b8 – cot \u03b8) ( cosec \u03b8 + cot \u03b8) = 1$
\nGiven cosec \u03b8 – cot \u03b8 = 3
\n$ \\therefore 3 ( cosec \u03b8 + cot \u03b8) = 1 $
\ncosec \u03b8 + cot \u03b8 = $\\Large\\frac{1}{3} $<\/p>\n

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

We know that,
\n$ \\Large\\frac{sinA+sinB}{cosA-cosB}$ = $ \\Large\\frac{2sin(\\frac{A+B}{2})cos(\\frac{A-B}{2})}{-2sin(\\frac{A+B}{2})sin(\\frac{A-B}{2})}$<\/p>\n

= -cot($\\frac{A-B}{2}$)<\/p>\n

Here, A = 75; B = 15<\/p>\n

Therefore,
\n$ \\Large\\frac{sin 75 + sin 15}{cos 75 – cos 15}$<\/p>\n

= -cot($\\Large\\frac{75 – 15}{2}$)<\/p>\n

= -cot($\\Large\\frac{60}{2}$)<\/p>\n

= -cot 30<\/p>\n

= $- \\sqrt{3}$<\/p>\n

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

Given,<\/p>\n

$\\cos \u03b8 = \\Large\\frac{24}{25}$ then $\\sin\u03b8 = \\Large\\frac{7}{25}$ &\u00a0$\\tan\u03b8 = \\Large\\frac{7}{24}$<\/p>\n

$\\tan(\u03b8) = \\Large\\frac{2\\tan(\u03b8\/2)}{1-\\tan^{2}(\u03b8\/2)} $<\/p>\n

$ \\Large\\frac{7}{24}$ = $\\Large\\frac{2\\tan(\u03b8\/2)}{1-\\tan^{2}(\u03b8\/2)}\u00a0$<\/p>\n

On solving we get,<\/p>\n

$tan (\u03b8\/2) = \\Large\\frac{1}{7}$ $ or -7$<\/p>\n

Since,\u00a00 < \u03b8 < 90<\/p>\n

$\\tan(\\frac{\u03b8}{2}) = \\Large\\frac{1}{7}$<\/p>\n

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

Given $sin x = \\dfrac{20}{29}$<\/p>\n

\u21d2 $sin^2 x = \\dfrac{400}{841}$<\/p>\n

We know that $cos^2 x = 1-sin^2 x$<\/p>\n

$cos^2 x = 1-\\dfrac{400}{841}$<\/p>\n

$= \\dfrac{441}{841}$<\/p>\n

$cos x= \\dfrac{21}{29}$<\/p>\n

$sec x = \\dfrac{1}{cos x} = \\dfrac{29}{21}$<\/p>\n

$tan x = \\dfrac{sin x}{cos x} = \\dfrac{20}{21}$<\/p>\n

$sec x – tan x = \\dfrac{29}{21} – \\dfrac{20}{21} = \\dfrac{9}{21} = \\dfrac{3}{7}$<\/p>\n

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

Given $sin x = \\dfrac{20}{29}$
\n\u21d2 $sin^2 x = \\dfrac{400}{841}$<\/p>\n

We know that $cos^2 x = 1-sin^2 x$<\/p>\n

$cos^2 x = 1-\\dfrac{400}{841}$<\/p>\n

$= \\dfrac{441}{841}$
\n$cos x= \\dfrac{21}{29}$<\/p>\n

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

$= \\dfrac{21}{20}$<\/p>\n

$cosec x = \\dfrac{1}{sin x} = \\dfrac{29}{20}$<\/p>\n

$cosec x – cot x = \\dfrac{29}{20} – \\dfrac{21}{20} = \\dfrac{29-21}{20} = \\dfrac{8}{20} = \\dfrac{2}{5}$<\/p>\n

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

We can write $sin^2 89 as sin^2 (90-1) = cos^2 1$
\n$sin^2 88 = sin^2 (90-2) = cos^2 2$
\nIn the same way, $sin^2 46 = sin^2 (90-44) = cos^2 44$
\nHence, the given equation can be written as,
\n$sin^2 1+cos^2 1+sin^2 2+cos^2 2+…..sin^2 45$
\n$= 1+1+1+1\u2026.+\\dfrac{1}{2}$
\nHere, the number of 1\u2019s will be $44$ as there are $88$ terms in the form of $sin^2 x$ and $cos^2 x$.
\nHence, Required answer $= 44+\\dfrac{1}{2} = 44.5$<\/p>\n

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

$sec 1^\\circ.sec 2^\\circ.sec 3^\\circ\u2026..sec 179^\\circ = sec 1^\\circ.sec 2^\\circ.sec 3^\\circ…\u2026sec 90^\\circ…..sec 179^\\circ$
\nWe know that $sec 90^\\circ = \\infty$
\nTherefore, The given series multiplication becomes $\\infty$.<\/p>\n\n

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We hope this Trigonometry\u00a0 Questions for RRB NTPC Exam will be highly useful for your Preparation<\/p>\n","protected":false},"excerpt":{"rendered":"

Trigonometry Questions for RRB NTPC PDF Download RRB NTPC Trigonometry Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Take a free mock test for RRB NTPC Download RRB NTPC Previous Papers PDF Question 1:\u00a0$\\frac{2sin\u03b8}{1+cos\u03b8}$ = ? a)\u00a02cot\u03b8\/2 b)\u00a0cot\u03b8\/2 […]<\/p>\n","protected":false},"author":41,"featured_media":28183,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603],"tags":[489,491,1647,1619],"class_list":{"0":"post-28181","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-ntpc","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-mocks","14":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":28183,"alt_text":"Trigonometry Questions for RRB NTPC PDF","caption":"Trigonometry Questions for RRB NTPC PDF","description":"Trigonometry Questions for RRB NTPC 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