{"id":40708,"date":"2020-02-24T18:43:40","date_gmt":"2020-02-24T13:13:40","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40708"},"modified":"2020-02-24T18:43:41","modified_gmt":"2020-02-24T13:13:41","slug":"trigonometry-questions-for-ssc-chsl-set-3-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/trigonometry-questions-for-ssc-chsl-set-3-pdf\/","title":{"rendered":"Trigonometry Questions for SSC CHSL set-3 PDF"},"content":{"rendered":"

Trigonometry Questions for SSC CHSL set-3 PDF<\/strong><\/span><\/h1>\n

Download SSC CHSL Trigonometry Questions with answers set-3 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exams.<\/p>\n

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Question 1:\u00a0<\/b>If $sin \u03b8 + sin^{2} \u03b8 = 1$ then $cos^2 \u03b8 + cos^4 \u03b8$ is equal to<\/p>\n

a)\u00a0None<\/p>\n

b)\u00a01<\/p>\n

c)\u00a0$Sin \u03b8 \/cos^{2} \u03b8$<\/p>\n

d)\u00a0$cos^{2}\u03b8 \/ Sin \u03b8$<\/p>\n

Question 2:\u00a0<\/b>The numerical value of $\\frac{cos^2 45\\circ}{sin^2 60\\circ}+\\frac{cos^2 60\\circ}{sin^2 45\\circ}-\\frac{tan^230\\circ}{cot^245\\circ}-\\frac{sin^230\\circ}{cot^230\\circ}$ is<\/p>\n

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

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

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

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

Question 3:\u00a0<\/b>The value of tan1\u00b0tan2\u00b0tan3\u00b0 \u2026\u2026\u2026\u2026\u2026tan89\u00b0 is<\/p>\n

a)\u00a01<\/p>\n

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

c)\u00a00<\/p>\n

d)\u00a0None of the options<\/p>\n

Question 4:\u00a0<\/b>If $\\frac{\\cos\\alpha}{\\sin\\beta}=n$ and $\\frac{\\cos\\alpha}{\\cos\\beta}=m$ then the value of $\\cos^{2} \\beta$ is <\/span><\/span><\/p>\n

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

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

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

d)\u00a00<\/p>\n

Question 5:\u00a0<\/b>If 0\u00b0 \u2264 A \u2264 90\u00b0, the simplified form of the given expression sin A cos A (tan A – cot A) is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a01 – 2 $sin^2$ A<\/p>\n

c)\u00a02 $sin^2$ A – 1<\/p>\n

d)\u00a01 – cos A<\/p>\n

Question 6:\u00a0<\/b>If \u03b8 is an acute angle and $\\tan^2\\theta+\\frac{1}{\\tan^2\\theta}=2$ then the value of \u03b8 is :<\/p>\n

a)\u00a060\u00b0<\/p>\n

b)\u00a045\u00b0<\/p>\n

c)\u00a015\u00b0<\/p>\n

d)\u00a030\u00b0<\/p>\n

Question 7:\u00a0<\/b>If tan \u03b8 + cot \u03b8 = 5, then $tan^2 \u03b8 + cot^2 \u03b8$ is<\/p>\n

a)\u00a023<\/p>\n

b)\u00a025<\/p>\n

c)\u00a026<\/p>\n

d)\u00a024<\/p>\n

Question 8:\u00a0<\/b>A person of height 6ft. wants to pluck a fruit which is on a 26\/3 ft. high tree. If the person is standing 8\/\u221a3 ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit ?<\/p>\n

a)\u00a075\u00b0<\/p>\n

b)\u00a030\u00b0<\/p>\n

c)\u00a045\u00b0<\/p>\n

d)\u00a060\u00b0<\/p>\n

Question 9:\u00a0<\/b>The value of $sin^{2}$ 22\u00b0 + $sin^{2}$ 68\u00b0 + $cot^{2}$ 30\u00b0 is<\/p>\n

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

b)\u00a0$3$<\/p>\n

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

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

Question 10:\u00a0<\/b>The minimum value of $2sin^{2}$ \u03b8 + $3cos^{2}$ \u03b8 is<\/p>\n

a)\u00a03<\/p>\n

b)\u00a04<\/p>\n

c)\u00a02<\/p>\n

d)\u00a01<\/p>\n

Question 11:\u00a0<\/b>If \u03b8 be acute angle and tan (4\u03b8 – 50\u00b0) = cot(50\u00b0 – \u03b8), then the value of 9 in degrees is:<\/p>\n

a)\u00a020<\/p>\n

b)\u00a050<\/p>\n

c)\u00a040<\/p>\n

d)\u00a030<\/p>\n

Question 12:\u00a0<\/b>If sec\u03b8 + tan\u03b8 = p, (p \u2260 0) then sec\u03b8 is equal to<\/p>\n

a)\u00a0$[p-\\frac{1}{p}], p\\neq 0$<\/p>\n

b)\u00a0$2[p-\\frac{1}{p}], p\\neq 0$<\/p>\n

c)\u00a0$[p+\\frac{1}{p}], p\\neq 0$<\/p>\n

d)\u00a0$\\frac{1}{2}[p+\\frac{1}{p}], p\\neq 0$<\/p>\n

Question 13:\u00a0<\/b>The value of $sin^{2} 30^{\\circ} cos^{2} 45^{\\circ}$ + $5tan^{2} 30^{\\circ}$ + $\\frac{3}{2} sin^{2} 90^{\\circ}$ – $3 cos^{2} 90^{\\circ}$ is<\/p>\n

a)\u00a0$3\\frac{7}{24}$<\/p>\n

b)\u00a0$3\\frac{3}{24}$<\/p>\n

c)\u00a0$3\\frac{1}{24}$<\/p>\n

d)\u00a0$3\\frac{5}{24}$<\/p>\n

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Question 14:\u00a0<\/b>If $cos^{2} \u03b8 – sin^{2} \u03b8 = \\frac{1}{3}$ , where 0 \u2264 \u03b8 \u2264 \u03c0\/2 then the value of $cos^{4} \u03b8 – sin^{4} \u03b8$ is<\/p>\n

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

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

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

d)\u00a0$\\frac{2}{9}$<\/p>\n

Question 15:\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

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

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

Expression : $sin\\theta + sin^2\\theta = 1$<\/p>\n

=> $sin\\theta = 1 – sin^2\\theta$<\/p>\n

=> $sin\\theta = cos^2\\theta$<\/p>\n

To find : $cos^2\\theta + cos^4\\theta$<\/p>\n

= $cos^2\\theta + (cos^2\\theta)^2$<\/p>\n

= $cos^2\\theta + sin^2\\theta$<\/p>\n

= 1<\/p>\n

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

Expression : $\\frac{cos^2 45}{sin^2 60}+\\frac{cos^2 60}{sin^2 45}-\\frac{tan^230}{cot^245}-\\frac{sin^230}{cot^230}$<\/p>\n

= $\\frac{(\\frac{1}{\\sqrt{2}})^2}{(\\frac{\\sqrt{3}}{2})^2} + \\frac{(\\frac{1}{2})^2}{(\\frac{1}{\\sqrt{2}})^2} – \\frac{(\\frac{1}{\\sqrt{3}})^2}{(1)^2} – \\frac{(\\frac{1}{2})^2}{(\\sqrt{3})^2}$<\/p>\n

= $(\\frac{1}{2} \\times \\frac{4}{3}) + (\\frac{1}{4} \\times 2) – (\\frac{1}{3} \\times 1) – (\\frac{1}{4 \\times 3})$<\/p>\n

= $\\frac{2}{3} + \\frac{1}{2} – \\frac{1}{3} – \\frac{1}{12}$<\/p>\n

= $\\frac{9}{12} = \\frac{3}{4}$<\/p>\n

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

Expression : tan1\u00b0tan2\u00b0tan3\u00b0 \u2026\u2026\u2026\u2026\u2026tan88\u00b0tan89\u00b0<\/p>\n

$\\because$ $tan(90^{\\circ}-\\theta) = cot\\theta$<\/p>\n

=> tan 89\u00b0 = tan(90\u00b0-1) = cot 1\u00b0<\/p>\n

Similarly, tan 88\u00b0 = cot 2\u00b0 and so on till tan 46\u00b0 = cot 44\u00b0<\/p>\n

=> (tan1\u00b0tan2\u00b0tan3\u00b0…….tan45\u00b0……cot3\u00b0cot2\u00b0cot1\u00b0)<\/p>\n

Using, $tan\\theta cot\\theta$ = 1 and $tan45^{\\circ}$ = 1<\/p>\n

=> 1*1 = 1<\/p>\n

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

$\\frac{\\cos\\alpha}{\\sin\\beta}=n$<\/p>\n

=> $cos\\alpha = nsin\\beta$<\/p>\n

and $\\frac{\\cos\\alpha}{\\cos\\beta}=m$<\/p>\n

=> $cos\\alpha = mcos\\beta$<\/span><\/p>\n

Comparing above equations, we get : <\/span><\/p>\n

=> $nsin\\beta = mcos\\beta$<\/span><\/p>\n

Squaring both sides : <\/span><\/p>\n

=> $n^2sin^2\\beta = m^2cos^2\\beta$<\/span><\/p>\n

=> $n^2(1-cos^2\\beta) = m^2cos^2\\beta$<\/span><\/p>\n

=> $n^2 = (n^2+m^2)cos^2\\beta$<\/span><\/p>\n

=> $cos^2\\beta = \\frac{n^2}{m^2+n^2}$<\/span><\/p>\n

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

Expression : $sin A cos A (tan A – cot A)$<\/p>\n

= $sin A cos A (\\frac{sin A}{cos A} – \\frac{cos A}{sin A})$<\/p>\n

= $sin A cos A (\\frac{sin^2 A – cos^2 A}{sin A cos A})$<\/p>\n

= $sin^2 A – cos^2 A$<\/p>\n

= $sin^2 A – (1 – sin^2 A)$<\/p>\n

= $2sin^2 A – 1$<\/p>\n

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

Expression : $\\tan^2\\theta+\\frac{1}{\\tan^2\\theta}=2$<\/p>\n

=> $(tan\\theta + \\frac{1}{tan\\theta})^2 – 2 = 2$<\/p>\n

=> $(tan\\theta + \\frac{1}{tan\\theta})^2 = 4$<\/p>\n

=> $tan\\theta + \\frac{1}{tan\\theta} = 2$<\/p>\n

[It can’t be -2 as $\\theta$ is in 1st quadrant, and $tan\\theta$ is positive in 1st quadrant.]<\/p>\n

=> $tan^2\\theta + 1 = 2tan\\theta$<\/p>\n

=> $(tan\\theta – 1)^2 = 0$<\/p>\n

=> $tan\\theta = 1$<\/p>\n

=> $\\theta = 45^{\\circ}$<\/p>\n

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

Expression : $tan\\theta + cot\\theta = 5$<\/p>\n

Squaring both sides, we get :<\/p>\n

=> $tan^2\\theta + cot^2\\theta + 2tan\\theta cot\\theta = 25$<\/p>\n

We know that, $tan\\theta cot\\theta = 1$<\/p>\n

=> $tan^2\\theta + cot^2\\theta = 25-2 = 23$<\/p>\n

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

<\/p>\n

Height of person = CD = 6 ft<\/p>\n

Height of tree = AB = $\\frac{26}{3}$ ft<\/p>\n

Distance between them = BD = $\\frac{8}{\\sqrt{3}}$ ft<\/p>\n

To find : $\\angle$ACE = $\\theta$ = ?<\/p>\n

Solution : AE = AB – BE = $\\frac{26}{3}$ – 6<\/p>\n

=> AE = $\\frac{8}{3} ft$<\/p>\n

and BD = CE = $\\frac{8}{\\sqrt{3}}$ ft<\/p>\n

Now, in $\\triangle$AEC<\/p>\n

=> $tan\\theta$ = $\\frac{AE}{CE}$<\/p>\n

=> $tan\\theta$ = $\\frac{\\frac{8}{3}}{\\frac{8}{\\sqrt{3}}}$<\/span><\/p>\n

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

=> $\\theta$ = 30\u00b0<\/span><\/span><\/span><\/p>\n

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

Expression : $sin^{2}$ 22\u00b0 + $sin^{2}$ 68\u00b0 + $cot^{2}$ 30\u00b0<\/p>\n

We know that $sin(90^{\\circ}-\\theta) = cos\\theta$<\/p>\n

=> $sin 22^{\\circ} = sin (90^{\\circ}-68^{\\circ}) = cos 68^{\\circ}$<\/p>\n

=> $cos^2 68^{\\circ} + sin^2 68^{\\circ} + (\\sqrt{3})^2$<\/p>\n

=> 1 + 3 = 4<\/p>\n

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

Expression : $2sin^{2}$ \u03b8 + $3cos^{2}$ \u03b8<\/p>\n

We know that, $sin^2 \\theta = 1 – cos^2 \\theta$<\/p>\n

=> $2 (1- cos^2 \\theta) + 3cos^2 \\theta$<\/p>\n

= $cos^2 \\theta + 2$<\/p>\n

Using the formula, $cos^2 \\theta = \\frac{cos 2\\theta + 1}{2}$<\/p>\n

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

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

$\\because$ minimum value of $cos 2\\theta = -1$<\/p>\n

=> min value = $\\frac{5}{2} – \\frac{1}{2} = 2$<\/p>\n

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

tan(90-\u03b8)=cot\u03b8
\ntan(4\u03b8-50) = cot(90 – (4\u03b8-50))
\n=cot(140-4\u03b8)
\nGiven tan (4\u03b8 – 50\u00b0) = cot(50\u00b0 – \u03b8)
\ntan (4\u03b8 – 50\u00b0) = cot(140\u00b0 – 4\u03b8)
\ncot(50\u00b0 – \u03b8) = cot(140\u00b0 – 4\u03b8)
\n50\u00b0 – \u03b8 = 140\u00b0 – 4\u03b8
\n3\u03b8 = 90\u00b0
\n\u03b8 = 30\u00b0
\nOption D is the correct answer.<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>
\n<\/span><\/span><\/span><\/span><\/p>\n

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

=> $sec\\theta + tan\\theta = p$ ————–Eqn(1)<\/span><\/p>\n

$\\because$ $sec^2\\theta – tan^2\\theta = 1$<\/span><\/p>\n

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

=> $(sec\\theta – tan\\theta) = \\frac{1}{p}$ ————–Eqn(2)<\/span><\/p>\n

Adding eqns(1)&(2)<\/span><\/p>\n

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

=> $sec\\theta = \\frac{1}{2}[p + \\frac{1}{p}]$<\/span><\/p>\n

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

Expression : $sin^{2} 30^{\\circ} cos^{2} 45^{\\circ}$ + $5tan^{2} 30^{\\circ}$ + $\\frac{3}{2} sin^{2} 90^{\\circ}$ – $3 cos^{2} 90^{\\circ}$<\/p>\n

= $[(\\frac{1}{2})^2 * (\\frac{1}{\\sqrt{2}})^2]$ + $5(\\frac{1}{\\sqrt{3}})^2 + \\frac{3}{2}(1^2) – 3(0)^2$<\/p>\n

= $ (\\frac{1}{4}*\\frac{1}{2}) + \\frac{5}{3} + \\frac{3}{2}$<\/p>\n

= $ \\frac{3+40+36}{24}$<\/p>\n

= $\\frac{79}{24} = 3\\frac{7}{24}$<\/p>\n

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

Expression : $cos^{2} \u03b8 – sin^{2} \u03b8 = \\frac{1}{3}$<\/p>\n

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

Adding the above two equations, we get :<\/p>\n

=> $2cos^2 \\theta = \\frac{4}{3}$<\/p>\n

=> $cos^2 \\theta = \\frac{2}{3}$<\/p>\n

Squaring both sides,<\/p>\n

=> $cos^4 \\theta = \\frac{4}{9}$<\/p>\n

Similarly, subtracting those two equations, we get :<\/p>\n

=> $sin^2 \\theta = \\frac{1}{3}$<\/p>\n

=> $sin^4 \\theta = \\frac{1}{9}$<\/p>\n

Now, to find : $cos^{4} \u03b8 – sin^{4} \u03b8$<\/p>\n

= $\\frac{4}{9} – \\frac{1}{9}$<\/p>\n

= $\\frac{3}{9} = \\frac{1}{3}$<\/p>\n

15)\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

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We hope this Trigonometry Questions of set-3 pdf for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Trigonometry Questions for SSC CHSL set-3 PDF Download SSC CHSL Trigonometry Questions with answers set-3 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exams. 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":49,"featured_media":40713,"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":[358,3658],"class_list":{"0":"post-40708","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-ssc-chsl","18":"tag-trigonometry-questions"},"better_featured_image":{"id":40713,"alt_text":"Trigonometry Questions for SSC CHSL set-3 PDF","caption":"Trigonometry Questions 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