{"id":8412,"date":"2024-11-30T09:45:23","date_gmt":"2024-11-30T04:15:23","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=8412"},"modified":"2025-12-10T17:11:27","modified_gmt":"2025-12-10T11:41:27","slug":"ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-2\/","title":{"rendered":"NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions – Ex 3.2"},"content":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0<\/span><\/strong><\/span><\/h2>\n

The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 3 Trigonometric Functions <\/strong>Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter 3 (Trigonometric Functions)\u00a0<\/span><\/strong><\/span><\/h2>\n\n\n\n
Exercise – 3.2<\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Find the values of other five trigonometric functions in Exercises 1 to 5. <\/strong><\/span><\/p>\n

1. cos x = -1\/2, x lies in third quadrant.
\n<\/strong>Solution –\u00a0<\/strong>Since cos x = (-1\/2)\u00a0 <\/span>
\nWe have sec x = 1\/cos x = -2 <\/span>
\nNow sin2<\/sup>\u00a0x + cos2<\/sup> x = 1 <\/span>
\ni.e., sin2\u00a0<\/sup>x = 1 \u2013 cos2<\/sup> x\u00a0 <\/span>
\nor sin2\u00a0<\/sup>x = 1 \u2013 (1\/4) = (3\/4)\u00a0 <\/span>
\nHence sin x = \u00b1(\u221a3\/2)\u00a0 <\/span>
\nSince x lies in third quadrant, sin x is negative. <\/span>
\nTherefore sin x = (\u2013\u221a3\/2)\u00a0 <\/span>
\nWhich also gives <\/span>
\ncosec x = 1\/sin x = (\u20132\/\u221a3)\u00a0 <\/span>
\nFurther, we have <\/span>
\ntan x = sin x\/cos x <\/span>
\n= (\u2013\u221a3\/2)\/(-1\/2) <\/span>
\n= \u221a3\u00a0 <\/span>
\nand <\/span>
\ncot x = 1\/tanx = (1\/\u221a3)\u00a0<\/span><\/p>\n

2. sin x = 3\/5, x lies in second quadrant.
\n<\/strong>Solution –\u00a0 <\/strong>Since sin x = (3\/5)\u00a0 <\/span>
\nwe have cosec x = 1\/sin x = (5\/3)\u00a0 <\/span>
\nNow sin2<\/sup>\u00a0x + cos2<\/sup> x = 1 <\/span>
\ni.e., cos2<\/sup>\u00a0x = 1 \u2013 sin2<\/sup> x <\/span>
\nor cos2 \u00a0<\/sup>x = 1 \u2013 (3\/5)\u00a0 <\/span>
\n= 1 \u2013 (9\/25)\u00a0 <\/span>
\n= (16\/25)\u00a0 <\/span>
\nHence cos x = \u00b1(4\/5)\u00a0 <\/span>
\nSince x lies in second quadrant, cos \u00a0x is negative.\u00a0 <\/span>
\nTherefore <\/span>
\ncos x = (\u20134\/5)\u00a0 <\/span>
\nwhich also gives <\/span>
\nsec x = 1\/cos x= (-5\/4)\u00a0 <\/span>
\nFurther, we have <\/span>
\ntan x = \u00a0sin x\/cos x <\/span>
\n= \u00a0(3\/5)\/(-4\/5)\u00a0 <\/span>
\n= \u00a0(-3\/4)\u00a0 <\/span>
\nand <\/span>
\ncot x = 1\/tan x = (-4\/3)<\/span><\/p>\n

3. cot x = 3\/4, x lies in third quadrant.
\n<\/strong>Solution –\u00a0<\/strong>Since cot x = (3\/4)\u00a0 <\/span>
\nwe have tan x = 1 \/ cot x = (4\/3)\u00a0 <\/span>
\nNow sec2<\/sup>\u00a0x = 1 + tan2\u00a0<\/sup>x <\/span>
\n= 1 + (16\/9) <\/span>
\n= (25\/9)\u00a0 <\/span>
\nHence sec x = \u00b1(5\/3)\u00a0 <\/span>
\nSince x lies in third quadrant, sec x \u00a0will be negative. Therefore <\/span>
\nsec x = (-5\/3)\u00a0 <\/span>
\nwhich also gives <\/span>
\ncos x = (-3\/5)\u00a0 <\/span>
\nFurther, we have <\/span>
\nsin x = tan x \u00a0* cos x <\/span>
\n= (4\/3) \u00d7 (-3\/5) <\/span>
\n= (-4\/5)\u00a0 <\/span>
\nand <\/span>
\ncosec x = 1\/sin x <\/span>
\n= (-5\/4)\u00a0 <\/span><\/p>\n

4. sec x = 13\/5, x lies in fourth quadrant.
\n<\/strong>Solution –\u00a0 <\/strong>Since sec x = (13\/5) <\/span>
\nwe have cos x = 1\/secx = (5\/13)\u00a0 <\/span>
\nNow sin2\u00a0<\/sup>x + cos2<\/sup> x = 1 <\/span>
\ni.e., sin2<\/sup>\u00a0x = 1 \u2013 cos2<\/sup> x <\/span>
\nor <\/span>
\nsin2<\/sup>\u00a0x = 1 \u2013 (5\/13)2
\n<\/sup>= 1 \u2013 (25\/169)<\/span>
\n= 144\/169 <\/span>
\nHence sin x = \u00b1(12\/13) <\/span>
\nSince x lies in forth quadrant, sin x is negative. Therefore <\/span>
\nsin x = (\u201312\/13) <\/span>
\nwhich also gives <\/span>
\ncosec x = 1\/sin x = (-13\/12) <\/span>
\nFurther, we have <\/span>
\ntan x = \u00a0sin x\/cos x <\/span>
\n= \u00a0(-12\/13) \/ (5\/13) <\/span>
\n= \u00a0(-12\/5) <\/span>
\nand <\/span>
\ncot x = 1\/tan x <\/span>
\n= (-5\/12)\u00a0<\/span><\/p>\n

5. tan x = -5\/12, x lies in second quadrant.
\n<\/strong>Solution –\u00a0<\/strong>Since tan x = (-5\/12) <\/span>
\nwe have cot x = 1\/tan x = (-12\/5) <\/span>
\nNow sec2<\/sup>\u00a0x = 1 + tan2<\/sup> x <\/span>
\n= 1 + (25\/144) <\/span>
\n= 169\/144 <\/span>
\nHence sec x = \u00b1(13\/12) <\/span>
\nSince x lies in second quadrant, sec x will be negative. Therefore <\/span>
\nsec x = (-13\/12) <\/span>
\nwhich also gives <\/span>
\ncos x = 1\/sec x = (-12\/13) <\/span>
\nFurther, we have <\/span>
\nsin x = tan x \u00a0* cos x <\/span>
\n= (-5\/12) \u00d7 (-12\/13)\u00a0<\/span>
\n= (5\/13) <\/span>
\nand <\/span>
\ncosec x = 1\/sin x <\/span>
\n= (13\/5)\u00a0<\/span><\/p>\n

Find the values of the trigonometric functions in Exercises 6 to 10. <\/strong><\/span><\/p>\n

6. sin 765\u00b0 <\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>We known that the values of sin x repeats after an interval of 2\u03c0 or 360\u2218<\/sup>.\u00a0 <\/span>
\nSo, <\/span>
\nsin(765\u00b0) <\/span>
\n= sin(720\u00b0 + 45\u00b0) { taking nearest multiple of 360 } <\/span>
\n= sin(2 \u00d7 360\u00b0 + 45\u00b0)\u00a0 <\/span>
\n= sin(45\u00b0)\u00a0 <\/span>
\n= 1\/\u221a2 <\/span>
\nHence, sin(765\u00b0) = 1\/\u221a2 <\/strong><\/span><\/p>\n

7. cosec (\u20131410\u00b0)
\n<\/strong>Solution –\u00a0<\/strong>We known that the values of cosec \u00a0x \u00a0repeats after an interval of 2\u03c0 or 360\u2218<\/sup>.\u00a0 <\/span>
\nSo, cosec(-1410\u00b0)\u00a0 <\/span>
\n= \u2013 cosec(1410\u00b0)\u00a0 <\/span>
\n= \u2013 cosec(1440\u00b0 \u2013 30\u00b0) \u00a0 \u00a0{ taking nearest multiple of 360 } <\/span>
\n= \u2013 cosec(4 \u00d7 360\u00b0 \u2013 30\u00b0)\u00a0 <\/span>
\n= cosec(30\u00b0)\u00a0 <\/span>
\n= 2 <\/span>
\nHence, cosec(\u20131410\u00b0) = 2.<\/strong><\/span><\/p>\n

8. tan(19\u03c0\/3)
\n<\/strong>Solution – <\/strong>We known that the values of tan x \u00a0repeats after an interval of \u03c0 or 180\u2218<\/sup>. <\/span>
\nSo, <\/span>
\ntan(19\u03c0\/3)\u00a0 <\/span>
\n\u00a0= tan(18\u03c0\/3 + \u03c0\/3) { breaking into nearest integer }\u00a0 <\/span>
\n\u00a0= tan(6\u03c0 + \u03c0\/3)<\/span>
\n\u00a0= tan(\u03c0\/3)\u00a0 <\/span>
\n\u00a0= tan(60\u00b0)\u00a0 <\/span>
\n\u00a0= \u221a3 <\/span>
\nHence, tan(19\u03c0\/3) = \u221a3.\u00a0<\/strong><\/span><\/p>\n

9. sin(\u201311\u03c0\/3)
\n<\/strong>Solution – <\/strong>We known that the values of sin x \u00a0repeats after an interval of 2\u03c0 or 360\u2218<\/sup>. <\/span>
\nSo, sin(-11\u03c0\/3) <\/span>
\n= -sin(11\u03c0\/3)\u00a0<\/span>
\n= -sin(12\u03c0\/3 \u2013 \u03c0\/3) { breaking nearest multiple of 2\u03c0 divisible by 3}<\/span>
\n= -sin(4\u03c0 \u2013 \u03c0\/3)<\/span>
\n= -sin(-\u03c0\/3)\u00a0<\/span>
\n= -[-sin(\u03c0\/3)]\u00a0<\/span>
\n= sin(\u03c0\/3)\u00a0<\/span>
\n= sin(60\u00b0)<\/span>
\n= \u221a3\/2 <\/span>
\nHence, sin(-11\u03c0\/3) = \u221a3\/2. <\/strong><\/span><\/p>\n

10. cot(\u201315\u03c0\/4)
\n<\/strong>Solution –\u00a0<\/strong>We known that the values of cot x repeats after an interval of \u03c0 or 180\u00b0.\u00a0 <\/span>
\nSo, cot(-15\u03c0\/4) <\/span>
\n= -cot(15\u03c0\/4) <\/span>
\n= -cot(16\u03c0\/4 \u2013 \u03c0\/4) { breaking into nearest multiple of \u03c0 \u00a0divisible by 4 } <\/span>
\n= -cot(4\u03c0 \u2013 \u03c0\/4) <\/span>
\n= -cot(-\u03c0\/4) <\/span>
\n= -[-cot(\u03c0\/4)]\u00a0 <\/span>
\n= cot(45\u00b0) <\/span>
\n= 1 <\/span>
\nHence, cot(-15\u03c0\/4) = 1.<\/strong><\/span><\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0 The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 3 Trigonometric Functions Exercise 3.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 3 (Trigonometric Functions)\u00a0 Exercise – 3.2 Find the values of other five trigonometric functions in Exercises 1 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1674],"tags":[1675,1679,1692,1694,1678,1680,1681],"class_list":["post-8412","post","type-post","status-publish","format-standard","hentry","category-class-11-maths","tag-class-11-ncert-mathematics-solutions","tag-class-11-ncert-solutions","tag-ncert-class-11-mathematics-chapter-3-trigonometric-functions-solutions","tag-ncert-class-11-mathematics-exercise-3-2-solutions","tag-ncert-class-11-mathematics-solutions","tag-ncert-solutions-class-11-mathematics","tag-ncert-solutions-class-11-maths"],"yoast_head":"\nNCERT Class 11 Maths Solutions - 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