{"id":40337,"date":"2020-01-31T14:43:52","date_gmt":"2020-01-31T09:13:52","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40337"},"modified":"2020-01-31T16:02:37","modified_gmt":"2020-01-31T10:32:37","slug":"trigonometry-questions-for-ssc-chsl-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/trigonometry-questions-for-ssc-chsl-set-2-pdf\/","title":{"rendered":"Trigonometry Questions for SSC CHSL Set-2 PDF"},"content":{"rendered":"

Trigonometry Questions for SSC CHSL Set-2 PDF<\/strong><\/span><\/h1>\n

Download SSC CHSL Trigonometry\u00a0 Questions with answers Set-2\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Questions for SSC Exam<\/p>\n

Download Trigonometry Questions Set-2 PDF<\/a><\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/a><\/p>\n

Take a free mock test for SSC CHSL<\/a><\/p>\n

Download SSC CHSL Previous Papers<\/a><\/p>\n

More SSC CHSL Important Questions and Answers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>In \u0394ABC, \u2220C = 90\u00b0 and AB = c, BC = a, CA = b; then the value of (cosec B – cos A) is<\/p>\n

a)\u00a0$\\frac{c^2}{ab}$<\/p>\n

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

c)\u00a0$\\frac{a^2}{bc}$<\/p>\n

d)\u00a0$\\frac{bc}{a^{2}}$<\/p>\n

Question 2:\u00a0<\/b>If tan\u03b8 = 1\/\u221a11 0 < \u03b8 < \u03c0\/2, then the value of $\\frac{cosec^{2}\\theta-\\sec^2\\theta}{cosec^2\\theta+\\sec^2\\theta}$<\/p>\n

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

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

c)\u00a0$\\frac{5}{6}$<\/p>\n

d)\u00a0$\\frac{6}{7}$<\/p>\n

Question 3:\u00a0<\/b>$\\frac{sin\\theta}{x} = \\frac{cos\\theta}{y}$, then $sin\\theta-cos\\theta$ is equal to<\/p>\n

a)\u00a0$x-y$<\/p>\n

b)\u00a0$x+y$<\/p>\n

c)\u00a0$\\frac{x-y}{\\sqrt{x^{2}+y^{2}}}$<\/p>\n

d)\u00a0$\\frac{y-x}{\\sqrt{x^{2}+y^{2}}}$<\/p>\n

Question 4:\u00a0<\/b>If x sin 60\u00b0.tan 30\u00b0 = sec 60\u00b0.cot 45\u00b0, then the value of x is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a02\u221a3<\/p>\n

c)\u00a04<\/p>\n

d)\u00a04\u221a3<\/p>\n

Question 5:\u00a0<\/b>A sphere and a hemisphere have the same radius. Then the ratio of their respective total surface areas is<\/p>\n

a)\u00a02 : 1<\/p>\n

b)\u00a01 : 2<\/p>\n

c)\u00a04 : 3<\/p>\n

d)\u00a03 : 4<\/p>\n

SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n\n

SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n

Question 6:\u00a0<\/b>If x*sin^3 \u03b8 + y *cos^3 \u03b8 = sin\u03b8 * cos\u03b8 \u2260 0 and x sin\u03b8 – y cos\u03b8 = 0, then value of (x^2 + y^2 ) is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a0sin\u03b8 – cos\u03b8<\/p>\n

c)\u00a0sin\u03b8 + cos\u03b8<\/p>\n

d)\u00a00<\/p>\n

Question 7:\u00a0<\/b>If 2 – cos^2 \u03b8 = 3 sin \u03b8 cos \u03b8, sin \u03b8 \u2260 cos \u03b8 then tan \u03b8 is<\/p>\n

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

b)\u00a00<\/p>\n

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

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

Question 8:\u00a0<\/b>If $tan\u03b8 = \\frac{a}{b}$, then $\\frac{a sin \\theta + b cos \\theta}{a sin \\theta – b cos\\theta}$ is<\/p>\n

a)\u00a0$\\frac{a}{a^2+b^2}$<\/p>\n

b)\u00a0$\\frac{b}{a^2+b^2}$<\/p>\n

c)\u00a0$\\frac{a^2-b^2}{a^2+b^2}$<\/p>\n

d)\u00a0$\\frac{a^2 + b^2}{a^2-b^2}$<\/p>\n

Question 9:\u00a0<\/b>If $sin \u03b8 + cos \u03b8 = \\sqrt{2} sin (90^{\\circ} – \u03b8)$, then cot \u03b8 is equal to<\/p>\n

a)\u00a0$\\sqrt{2}+1$<\/p>\n

b)\u00a0$\\frac{1}{\\sqrt{2}}+2$<\/p>\n

c)\u00a0$\\sqrt{20}-1$<\/p>\n

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

Question 10:\u00a0<\/b>What is the value of cosec 11\u03c0\/6?<\/p>\n

a)\u00a0-2\/\u221a3<\/p>\n

b)\u00a0\u221a2<\/p>\n

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

d)\u00a0-\u221a2<\/p>\n

SSC CHSL FREE MOCK TEST<\/a><\/p>\n

FREE SSC STUDY MATERIAL<\/a><\/p>\n

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

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

<\/p>\n

In $\\triangle$$ABC, AB^2 = AC^2 + BC^2$<\/p>\n

=> $c^2 = a^2 + b^2 => c^2 – b^2 = a^2$<\/p>\n

$cosecB = \\frac{AB}{BC} = \\frac{c}{b}$<\/p>\n

$cosA = \\frac{AC}{AB} = \\frac{b}{c}$<\/p>\n

$\\therefore cosecB – cosA = \\frac{c}{b} – \\frac{b}{c}$<\/p>\n

= $\\frac{c^2-b^2}{bc} = \\frac{a^2}{bc}$<\/p>\n

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

Expression : $tan\\theta = \\frac{1}{\\sqrt{11}}$<\/p>\n

We know that, $sec\\theta = \\sqrt{1 + tan^2 \\theta}$<\/p>\n

=> $sec\\theta = \\sqrt{1 + \\frac{1}{11}} = \\sqrt{\\frac{12}{11}}$<\/p>\n

Now, $cosec\\theta = \\frac{sec\\theta}{tan\\theta}$<\/p>\n

=> $cosec\\theta = \\sqrt{12}$<\/p>\n

To find : $\\frac{cosec^{2}\\theta-\\sec^2\\theta}{cosec^2\\theta+\\sec^2\\theta}$<\/p>\n

= $\\frac{12 – \\frac{12}{11}}{12 + \\frac{12}{11}}$<\/p>\n

= $\\frac{1 – \\frac{1}{11}}{1 + \\frac{1}{11}}$<\/p>\n

= $\\frac{10}{12} = \\frac{5}{6}$<\/p>\n

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

$\\frac{sin\\theta}{x} = \\frac{cos\\theta}{y}$<\/p>\n

Reaaranging the given data , we get<\/p>\n

${tan\\theta}$ = $\\frac{x}{y}$<\/p>\n

Now taking ${cos\\theta}$ common from $sin\\theta-cos\\theta$,we get<\/p>\n

= $cos\\theta{(tan\\theta) – 1}$…………(1)<\/p>\n

Imagine a right angle triangle<\/p>\n

<\/p>\n

From this triangle , we can calculate values of $cos\\theta$ and $tan\\theta$ and hence putting the values in equation 1<\/p>\n

we get = $\\frac{y}{\\surd (x^2 + y^2)}$ ($\\frac{x}{y} $- 1)<\/p>\n

= $\\frac{x-y}{\\sqrt{x^{2}+y^{2}}}$<\/p>\n

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

sin 60\u00b0 = $\\frac{\\sqrt(3)}{2}$<\/p>\n

tan 30 = $\\frac{1}{\\sqrt(3)}$<\/span><\/p>\n

sec 60 = 2<\/span><\/span><\/p>\n

cot 45 = 1<\/span><\/span><\/p>\n

x $\\frac{\\sqrt(3)}{2}$ $\\frac{1}{\\sqrt(3)}$ = $(2\\times1)$<\/p>\n

x = 4<\/p>\n

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

Let the common radius be “r” units<\/p>\n

area of sphere = 4$\\pi r^2$<\/p>\n

area of hemi-sphere = 2$\\pi r^2$ + $\\pi r^2$ = 3$\\pi r^2$<\/p>\n

$\\frac{\\text{Area of Sphere}}{\\text{Area of Hemisphere}}$ = $\\frac{4\\pi r^2}{ 3\\pi r^2}$<\/p>\n

= $\\frac{4}{3}$<\/p>\n

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

Given, xsin\u03b8 -ycos\u03b8 = 0
\ny<\/span>=<\/span>x<\/span>s<\/span>i<\/span>n<\/span>\u03b8<\/span><\/span><\/span>c<\/span>o<\/span>s<\/span>\u03b8<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

\u21d2y=xsin\u03b8cos\u03b8<\/p>\n

Given the expression:
\nxsin$^3$\u03b8 + ycos$^3$\u03b8 = sin\u03b8cos\u03b8
\nReplacing the value of y we get,
\nx<\/span>s<\/span>i<\/span>n<\/span>$^3$<\/span><\/span><\/span><\/span>\u03b8<\/span>+(<\/span>x<\/span>s<\/span>i<\/span>n<\/span>\u03b8\/<\/span><\/span><\/span>c<\/span>o<\/span>s<\/span>\u03b8)<\/span><\/span><\/span><\/span>\u00d7<\/span>c<\/span>o<\/span>s<\/span>$^3$<\/span><\/span><\/span><\/span>\u03b8<\/span>=<\/span>s<\/span>i<\/span>n<\/span>\u03b8<\/span>c<\/span>o<\/span>s<\/span>\u03b8
\n<\/span>x sin3\u03b8 + x sin\u03b8cos$^2$\u03b8 = sin\u03b8cos\u03b8
\nx sin\u03b8 \u00d7 (sin2\u03b8 + cos2\u03b8) = sin\u03b8cos\u03b8
\nWe know the identity: sin2\u03b8 + cos2\u03b8 = 1
\nx = cos\u03b8
\ny<\/span>=<\/span>x<\/span>s<\/span>i<\/span>n<\/span>\u03b8\/<\/span><\/span><\/span>c<\/span>o<\/span>s<\/span>\u03b8
\n<\/span><\/span><\/span><\/span>y = sin\u03b8
\nHence,
\nx$^2$ + y$^2$ = cos$^2$\u03b8 + sin$^2$\u03b8
\n= 1
\nOption A is the correct answer.<\/p>\n

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

$2-cos^2 \\theta = 1+1-cos^2 \\theta = 1+sin^2 \\theta$
\nDividing the LHS and RHS by $cos^2\\theta$
\n$1+sin^2 \\theta = sec^2 \\theta + tan^2 \\theta$
\n$3sin \\theta cos \\theta = 3 tan \\theta$
\n$sec ^2 \\theta + tan^2 \\theta = 3tan \\theta$
\n$sec ^2 \\theta = 1+tan^2 \\theta$
\n$1+tan^2 \\theta+tan^2 \\theta = 3tan \\theta$
\n$1+2tan^2 \\theta = 3tan \\theta$
\n$2tan^2 \\theta – 3tan \\theta + 1 =0$
\nlet x=$tan \\theta$
\nThe equation becomes
\n$2x^2 + 3x + 1=0$
\nOn solving for x we get x=$1$ and x=$\\frac{1}{2}$
\nOption A is the correct answer.<\/span><\/p>\n

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

To find : $\\frac{a sin \\theta + b cos \\theta}{a sin \\theta – b cos\\theta}$<\/p>\n

Dividing numerator and denominator by $cos \\theta$, we get :<\/p>\n

= $\\frac{a tan \\theta + b}{a tan \\theta – b}$<\/p>\n

Also, it is given that $tan\u03b8 = \\frac{a}{b}$<\/p>\n

= $\\frac{a \\times \\frac{a}{b} + b}{a \\times \\frac{a}{b} – b}$<\/p>\n

= $\\frac{\\frac{a^2 + b^2}{b}}{\\frac{a^2 – b^2}{b}}$<\/p>\n

= $\\frac{a^2 + b^2}{a^2 – b^2}$<\/p>\n

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

Expression : $sin \u03b8 + cos \u03b8 = \\sqrt{2} sin (90^{\\circ} – \u03b8)$<\/p>\n

=> $sin \\theta + cos \\theta = \\sqrt{2} cos \\theta$<\/p>\n

=> $sin \\theta = cos \\theta (\\sqrt{2} – 1)$<\/p>\n

=> $\\frac{cos \\theta}{sin \\theta} = \\frac{1}{\\sqrt{2} – 1}$<\/p>\n

=> $cot \\theta = \\frac{1}{\\sqrt{2} – 1} \\times \\frac{\\sqrt{2} + 1}{\\sqrt{2} + 1}$<\/p>\n

=> $cot \\theta = \\sqrt{2} + 1$<\/p>\n

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

Expression\u00a0:\u00a0cosec 11\u03c0\/6<\/p>\n

= $cosec(2\\pi – \\frac{\\pi}{6})$<\/p>\n

= $-cosec(\\frac{\\pi}{6}) = -2$<\/p>\n

=> Ans – (C)<\/p>\n

DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n

SSC CHSL Important Q&A PDF<\/a><\/p>\n

We hope this Trigonometry Questions for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Trigonometry Questions for SSC CHSL Set-2 PDF Download SSC CHSL Trigonometry\u00a0 Questions with answers Set-2\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Questions for SSC Exam Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers More SSC CHSL Important Questions and Answers PDF Question […]<\/p>\n","protected":false},"author":42,"featured_media":40345,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[1254,169,125,9,504,378,1493,1459,1611,1741,1268,1441,1],"tags":[358,3583],"class_list":{"0":"post-40337","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-competitive-exams","8":"category-downloads","9":"category-featured","10":"category-ssc","11":"category-ssc-cgl","12":"category-ssc-chsl","13":"category-ssc-cpo","14":"category-ssc-gd","15":"category-ssc-je","16":"category-ssc-mts","17":"category-ssc-stenographer","18":"category-stenographer","19":"category-uncategorized","20":"tag-ssc-chsl","21":"tag-trigonometry-questions-for-ssc-chsl"},"better_featured_image":{"id":40345,"alt_text":"Trigonometry Questions for SSC CHSL Set-2 PDF","caption":"Trigonometry Questions for SSC CHSL Set-2 PDF\n","description":"Trigonometry Questions for SSC CHSL Set-2 PDF\n","media_type":"image","media_details":{"width":1200,"height":630,"file":"2020\/01\/fig-31-01-2020_08-02-17.jpg","sizes":{"thumbnail":{"file":"fig-31-01-2020_08-02-17-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-150x150.jpg"},"medium":{"file":"fig-31-01-2020_08-02-17-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-300x158.jpg"},"medium_large":{"file":"fig-31-01-2020_08-02-17-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-768x403.jpg"},"large":{"file":"fig-31-01-2020_08-02-17-1024x538.jpg","width":1024,"height":538,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-1024x538.jpg"},"tiny-lazy":{"file":"fig-31-01-2020_08-02-17-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-30x16.jpg"},"td_80x60":{"file":"fig-31-01-2020_08-02-17-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-80x60.jpg"},"td_100x70":{"file":"fig-31-01-2020_08-02-17-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-100x70.jpg"},"td_218x150":{"file":"fig-31-01-2020_08-02-17-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-218x150.jpg"},"td_265x198":{"file":"fig-31-01-2020_08-02-17-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-265x198.jpg"},"td_324x160":{"file":"fig-31-01-2020_08-02-17-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-324x160.jpg"},"td_324x235":{"file":"fig-31-01-2020_08-02-17-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-324x235.jpg"},"td_324x400":{"file":"fig-31-01-2020_08-02-17-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-324x400.jpg"},"td_356x220":{"file":"fig-31-01-2020_08-02-17-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-356x220.jpg"},"td_356x364":{"file":"fig-31-01-2020_08-02-17-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-356x364.jpg"},"td_533x261":{"file":"fig-31-01-2020_08-02-17-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-533x261.jpg"},"td_534x462":{"file":"fig-31-01-2020_08-02-17-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-534x462.jpg"},"td_696x0":{"file":"fig-31-01-2020_08-02-17-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-696x365.jpg"},"td_696x385":{"file":"fig-31-01-2020_08-02-17-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-696x385.jpg"},"td_741x486":{"file":"fig-31-01-2020_08-02-17-741x486.jpg","width":741,"height":486,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-741x486.jpg"},"td_1068x580":{"file":"fig-31-01-2020_08-02-17-1068x580.jpg","width":1068,"height":580,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-1068x580.jpg"},"td_1068x0":{"file":"fig-31-01-2020_08-02-17-1068x561.jpg","width":1068,"height":561,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-1068x561.jpg"},"td_0x420":{"file":"fig-31-01-2020_08-02-17-800x420.jpg","width":800,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17-800x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0","keywords":[]}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/01\/fig-31-01-2020_08-02-17.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n