{"id":40540,"date":"2020-03-03T10:10:03","date_gmt":"2020-03-03T04:40:03","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40540"},"modified":"2020-03-03T10:10:03","modified_gmt":"2020-03-03T04:40:03","slug":"ssc-questions-on-trigonometry-pdf-2","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/ssc-questions-on-trigonometry-pdf-2\/","title":{"rendered":"SSC Questions on Trigonometry PDF"},"content":{"rendered":"

SSC Questions on Trigonometry PDF<\/strong><\/span><\/h1>\n

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

Download SSC Questions on Trigonometry PDF<\/a><\/p>\n

Take a free SSC-CGL mock test<\/a><\/p>\n

Get 200 SSC mocks for just Rs. 249. Enroll here<\/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

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

Download SSC CGL Previous Papers PDF<\/a><\/p>\n

18000+ Questions – Free SSC Study Material<\/a><\/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

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

FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n

More SSC CGL Important Questions and Answers PDF<\/a><\/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

SSC CGL Previous Papers Download PDF<\/a><\/p>\n

SSC CGL Free Mock Test<\/a><\/p>\n

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

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

SSC Free Previous Papers App<\/a><\/p>\n

We hope this Trigonometry Questions for SSC\u00a0 Exams will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

SSC Questions on Trigonometry PDF Download SSC\u00a0 Questions on Trigonometry PDF based on previous papers very useful for SSC\u00a0 Exams. Top-10 Very Important Trigonometry Questions for SSC\u00a0 Exam. Question 1:\u00a0$5tan\\theta = 4$, then the value of\u00a0$(\\frac{5sin\\theta – 3cos\\theta}{5sin\\theta + 3cos\\theta})$ is a)\u00a0$\\frac{1}{7}$ b)\u00a0$\\frac{2}{7}$ c)\u00a0$\\frac{5}{7}$ d)\u00a0$\\frac{2}{5}$ Question 2:\u00a0The least value of $(4sec^2\\theta + 9cosec^2\\theta)$ is a)\u00a01 […]<\/p>\n","protected":false},"author":49,"featured_media":40940,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1441],"tags":[3638,165,462],"class_list":{"0":"post-40540","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-stenographer","17":"tag-questions-on-trigonometry","18":"tag-ssc","19":"tag-ssc-cgl"},"better_featured_image":{"id":40940,"alt_text":"SSC Questions on Trigonometry PDF","caption":"SSC Questions on Trigonometry PDF","description":"SSC Questions on Trigonometry PDF","media_type":"image","media_details":{"width":800,"height":420,"file":"2020\/03\/fig-01-03-2020_05-48-46.jpg","sizes":{"medium":{"file":"fig-01-03-2020_05-48-46-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-300x158.jpg"},"thumbnail":{"file":"fig-01-03-2020_05-48-46-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-150x150.jpg"},"medium_large":{"file":"fig-01-03-2020_05-48-46-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-768x403.jpg"},"tiny-lazy":{"file":"fig-01-03-2020_05-48-46-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-30x16.jpg"},"td_218x150":{"file":"fig-01-03-2020_05-48-46-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-218x150.jpg"},"td_324x400":{"file":"fig-01-03-2020_05-48-46-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-324x400.jpg"},"td_696x0":{"file":"fig-01-03-2020_05-48-46-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-696x365.jpg"},"td_80x60":{"file":"fig-01-03-2020_05-48-46-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-80x60.jpg"},"td_100x70":{"file":"fig-01-03-2020_05-48-46-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-100x70.jpg"},"td_265x198":{"file":"fig-01-03-2020_05-48-46-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-265x198.jpg"},"td_324x160":{"file":"fig-01-03-2020_05-48-46-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-324x160.jpg"},"td_324x235":{"file":"fig-01-03-2020_05-48-46-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-324x235.jpg"},"td_356x220":{"file":"fig-01-03-2020_05-48-46-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-356x220.jpg"},"td_356x364":{"file":"fig-01-03-2020_05-48-46-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-356x364.jpg"},"td_533x261":{"file":"fig-01-03-2020_05-48-46-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-533x261.jpg"},"td_534x462":{"file":"fig-01-03-2020_05-48-46-534x420.jpg","width":534,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-534x420.jpg"},"td_696x385":{"file":"fig-01-03-2020_05-48-46-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-696x385.jpg"},"td_741x486":{"file":"fig-01-03-2020_05-48-46-741x420.jpg","width":741,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46-741x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0"}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/03\/fig-01-03-2020_05-48-46.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n