{"id":25818,"date":"2019-03-07T15:55:24","date_gmt":"2019-03-07T10:25:24","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=25818"},"modified":"2020-01-31T16:02:34","modified_gmt":"2020-01-31T10:32:34","slug":"trigonometry-questions-for-ssc-chsl-exam-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/trigonometry-questions-for-ssc-chsl-exam-pdf\/","title":{"rendered":"Trigonometry Questions for SSC CHSL Exam PDF"},"content":{"rendered":"

Trigonometry Questions for SSC CHSL Exam PDF:<\/span><\/strong><\/h2>\n

SSC CHSL Trignometry Questions download PDF based on previous year question paper of SSC CHSL exam. 25 Very important Trignometry questions for SSC CHSL Exam.<\/p>\n

Download Trigonometry Questions for SSC CHSL Exam PDF<\/a><\/p>\n

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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

 <\/p>\n

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

a)\u00a00
\nb)\u00a0-1
\nc)\u00a01
\nd)\u00a02<\/p>\n

Question 2:\u00a0<\/b>If $cos^2x + (1 + sinx)^2 = 3$, then find the value of cosec x + sin(60+x), where 0\u00ba < x < 90\u00ba?<\/p>\n

a)\u00a03
\nb)\u00a02
\nc)\u00a0$2\\frac{1}{2}$
\nd)\u00a0$3\\frac{1}{2}$<\/p>\n

Question 3:\u00a0<\/b>If $x$ and $y$ are acute angles such that their sum is less than $90^{\\circ}$ . If $\\cos(2x-20^{\\circ})=\\sin(2y+20^{\\circ})$ , then the value of $\\tan(x+y)$ is<\/p>\n

a)\u00a0$0$
\nb)\u00a0$1$
\nc)\u00a0$\\frac{1}{\\sqrt{3}}$
\nd)\u00a0$\\sqrt{3}$<\/p>\n

Question 4:\u00a0<\/b>If x is an acute angle such that
\nTan (4x – 50\u00b0) = Cot ( 50\u00b0 – x), then the value of x would be?<\/p>\n

a)\u00a060
\nb)\u00a045
\nc)\u00a050
\nd)\u00a030<\/p>\n

Question 5:\u00a0<\/b>If cosec 9x= sec 9x (0 <x<10), what is the value of x?<\/p>\n

a)\u00a09\u00b0
\nb)\u00a03\u00b0
\nc)\u00a05\u00b0
\nd)\u00a06\u00b0<\/p>\n

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Question 6:\u00a0<\/b>If $\\sin x+1=\\frac{2}{\\sin x} $ then $\\cos ^3 x+ \\mathrm{cosec} ^{-3} x=?$<\/p>\n

a)\u00a00
\nb)\u00a0-1
\nc)\u00a01
\nd)\u00a02<\/p>\n

Question 7:\u00a0<\/b>If $\\frac{sin2x}{sinx} + cos2x + sin^2x = 0$, find the value of x, if $0 \\leq x \\leq 180$.<\/p>\n

a)\u00a045\u00ba
\nb)\u00a090\u00ba
\nc)\u00a0180\u00ba
\nd)\u00a0More than one value is possible<\/p>\n

Question 8:\u00a0<\/b>If $\\frac{sin^2 \\Theta}{4} = \\frac{cos^2 \\Theta}{9}$. Find the value of $tan^2 \\Theta – cot^2 \\Theta$.<\/p>\n

a)\u00a01\/6
\nb)\u00a013\/27
\nc)\u00a01\/18
\nd)\u00a0None of these<\/p>\n

Question 9:\u00a0<\/b>If $\\sec^4 x-\\tan^4 x=4+4sin^2 x$ then $ x=?$ (0<x<90\u00b0)<\/p>\n

a)\u00a045\u00b0
\nb)\u00a030\u00b0
\nc)\u00a075\u00b0
\nd)\u00a060\u00b0<\/p>\n

Question 10:\u00a0<\/b>Simplify the expression $\\sqrt{\\frac{1+cosx}{1-cosx}}$.<\/p>\n

a)\u00a0sec x + tan x
\nb)\u00a0sin x
\nc)\u00a0cosec x + cot x
\nd)\u00a0cos x<\/p>\n

 <\/p>\n

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Question 11:\u00a0<\/b>If $sinx = 1 – cos x$, find the value of cos 2x.<\/p>\n

a)\u00a00
\nb)\u00a01
\nc)\u00a0$\\frac{1}{2}$
\nd)\u00a0$\\frac{\\sqrt{3}}{2}$<\/p>\n

Question 12:\u00a0<\/b>If it is known that
\n$\\tan\\theta + \\cot\\theta = 2$, then find the value of $\\tan^3\\theta + \\cot^3\\theta$<\/p>\n

a)\u00a0$\\sqrt{5}$
\nb)\u00a02
\nc)\u00a0$1 + \\sqrt{3}$
\nd)\u00a0$2 + \\sqrt{3}$<\/p>\n

Question 13:\u00a0<\/b>If $\\sin x+\\frac{1}{\\sin x}=\\frac{5}{2}$ then $\\tan {x}+\\cot {x}=?$<\/p>\n

a)\u00a00
\nb)\u00a0$\\frac{2}{\\sqrt3}$
\nc)\u00a0$\\frac{4}{\\sqrt3}$
\nd)\u00a0$\\frac{\\sqrt3}{2}$<\/p>\n

Question 14:\u00a0<\/b>If $\\cos x+\\frac{1}{\\cos x}=2$ then $\\sin ^{5} x+\\sec ^{-5} x=?$<\/p>\n

a)\u00a0-1
\nb)\u00a00
\nc)\u00a01
\nd)\u00a02<\/p>\n

Question 15:\u00a0<\/b>If $\\tan x+\\frac{1}{\\tan x}=2$ then $\\sin ^{5} x+\\cos ^{5} x=?$<\/p>\n

a)\u00a0$\\frac{1}{4\\sqrt{2}}$
\nb)\u00a0$\\frac{1}{2\\sqrt{2}}$
\nc)\u00a0$\\frac{1}{\\sqrt{2}}$
\nd)\u00a0$\\frac{1}{8\\sqrt{2}}$<\/p>\n

FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n

Question 16:\u00a0<\/b>The maximum value of sin \u03b8 + cos \u03b8 is<\/p>\n

a)\u00a0$1$
\nb)\u00a0$\\sqrt{2}$
\nc)\u00a0$2$
\nd)\u00a0$3$<\/p>\n

Question 17:\u00a0<\/b>If $sin \\Theta + cosec \\Theta = 5$, Find the value of $sin^3 \\Theta + cosec ^3 \\Theta $.<\/p>\n

a)\u00a096
\nb)\u00a0115
\nc)\u00a0110
\nd)\u00a0None of these<\/p>\n

Question 18:\u00a0<\/b>The value of $cos^2 30^{\\circ} + sin^2 60^{\\circ} + tan^2 45^{\\circ} + sec^2 60^{\\circ} + cos0^{\\circ}$ is<\/p>\n

a)\u00a0$4\\frac{1}{2}$
\nb)\u00a0$5\\frac{1}{2}$
\nc)\u00a0$6\\frac{1}{2}$
\nd)\u00a0$7\\frac{1}{2}$<\/p>\n

Question 19:\u00a0<\/b>If $cos x + cos^{2} x = 1,$ then $sin^{8} x + 2 sin^{6} x + sin^{4}$ x is equal to<\/p>\n

a)\u00a00
\nb)\u00a03
\nc)\u00a02
\nd)\u00a01<\/p>\n

Question 20:\u00a0<\/b>If $x= p$ $cosec \u03b8$ and $y= q$ $cot \u03b8$, then the value of $\\frac{x^2}{p^2}-\\frac{y^2}{q^2}$ is<\/p>\n

a)\u00a0$sin^{2} \u03b8$
\nb)\u00a0$tan\u03b8$
\nc)\u00a0$1$
\nd)\u00a0$0$<\/p>\n

 <\/p>\n

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Question 21:\u00a0<\/b>Find the value of tan 4\u00b0 tan 43\u00b0 tan 47 tan 86\u00b0<\/p>\n

a)\u00a0$\\frac{2}{3}$
\nb)\u00a0$1$
\nc)\u00a0$\\frac{1}{2}$
\nd)\u00a0$2$<\/p>\n

Question 22:\u00a0<\/b>If $x cos\u03b8 – sin\u03b8 = 1,$ then $x^{2} – (1 +x^{2}) sin\u03b8$ equals<\/p>\n

a)\u00a02
\nb)\u00a01
\nc)\u00a0– 1
\nd)\u00a00<\/p>\n

Question 23:\u00a0<\/b>If $sin \u03b8 + sin^{2} \u03b8 = 1$ then $cos^2 \u03b8 + cos^4 \u03b8$ is equal to<\/p>\n

a)\u00a0None
\nb)\u00a01
\nc)\u00a0$Sin \u03b8 \/cos^{2} \u03b8$
\nd)\u00a0$cos^{2}\u03b8 \/ Sin \u03b8$<\/p>\n

Question 24:\u00a0<\/b>The value of tan1\u00b0tan2\u00b0tan3\u00b0 \u2026\u2026\u2026\u2026\u2026tan89\u00b0 is<\/p>\n

a)\u00a01
\nb)\u00a0-1
\nc)\u00a00
\nd)\u00a0None of the options<\/p>\n

Question 25:\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
\nb)\u00a01 – 2 $sin^2$ A
\nc)\u00a02 $sin^2$ A – 1
\nd)\u00a01 – cos A<\/p>\n

FREE MOCK TEST FOR SSC STENOGRAPHER<\/a><\/p>\n

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

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

$\\sin x+\\frac{1}{\\sin x}=2$
\n=>$\\sin^2 x+1=2\\sin x$
\n=>$(\\sin x-1)^2=0$
\n=>$\\sin x = 1$
\n=>$\\cos x = 0$
\nThe given expression would be,
\n$\\cos ^5 x+\\cot ^{5} x$
\n$\\cos ^5 x+\\frac{\\cos^5 x}{\\sin^5 x}$
\n$=0$<\/p>\n

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

$cos^2x + (1 + sinx)^2 = 3$
\n=> $cos^2x + 1 + sin^2x + 2sinx = 3$ (Since $ cos^2x + sin^2x = 1$)
\n=> $2 sin x = 1$ or sin x = \u00bd which means x=30 degrees [Solution in the first quadrant]
\n=> cosec x = 1\/sin x = 2 and sin (60+x) = sin 90 = 1
\n=> The value of the expression is 2+1 = 3<\/p>\n

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

Given that , $x+y<90^{\\circ} \\dots (1)$
\n$\\implies\\cos(2x-20^{\\circ})=\\cos(90^{\\circ}-(2y+20^{\\circ}))$
\nUsing the given property in $(1)$
\n$\\implies 2x-20^{\\circ}=70^{\\circ}-2y$
\n$\\implies 2x+2y=90^{\\circ}$
\n$\\implies x+y=45^{\\circ}$
\n$\\implies \\tan(x+y)=1 $
\nHence , the correct option is B<\/p>\n

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

We know that Tan ( 90 – x) = Cot x and Cot (90 – x) = Tanx
\nSo 4x – 50 = 90 – y
\nThen 50 – x will be y
\n=> y = 50 – x
\nx + y = 50
\nMoreover,
\n90 – y = 4x – 50
\n=> 4x + y = 140
\nSubtracting the two equations, we get
\n3x = 90
\n=> x = 30
\nHence the correct answer is option D.<\/p>\n

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

If cosec 9x= sec 9x,
\nThen $\\sin 9x=\\cos 9x$
\nIn the first quadrant cos and sin are equal when $9x =45$\u00b0
\nHence, x = 5\u00b0<\/p>\n

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

$\\sin x+1=\\frac{2}{\\sin x} $
\n=>$\\sin^2 x+\\sin x=2$
\n=>$(\\sin x-1)(\\sin x+2)=0$
\nAs $\\sin x$ can only take values between -1 and 1,
\n=>$\\sin x = 1$
\n=>$\\cos x = 0$
\nThe given expression would be,
\n$\\cos ^3 x+ \\mathrm{cosec}^{-3} x$
\n$\\cos ^3 x + \\sin^3 x$
\n$=1$<\/p>\n

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

$\\frac{sin2x}{sinx} + cos2x + sin^2x = 0$
\nWe know that $sin2x = 2 sinx cosx$ and $cos2x=cos^2x-sin^2x$
\n=> $2cosx + cos^2x = 0$
\n=> cos x =0 or cos x =2
\nNow, cos x =2 is not possible.
\n=> cos x = 0
\nIn the range, $0 \\leq x \\leq 180$, cos x =0 at only 90\u00ba
\nThus, B is the correct answer.<\/p>\n

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

$\\frac{sin^2 \\Theta}{4} = \\frac{cos^2 \\Theta}{9}$
\n==> $\\frac{sin^2 \\Theta}{cos^2 \\Theta}$ = $\\frac{4}{9}$
\n==> $tan^2 \\Theta = \\frac{4}{9}$
\n==> $cot^2 \\Theta = \\frac{9}{4}$
\n==> $tan^2 \\Theta – cot^2 \\Theta$ = 4\/9 – 9\/4 = -65\/36
\nSince there is no such option, the correct option to choose is D.<\/p>\n

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

$\\sec^4 x-\\tan^4 x=4+4sin^2 x$
\n$(sec^2 x-\\tan^2 x)( sec^2 x+\\tan^2 x)=4+4sin^2 x$
\n$(1)( sec^2 x+\\tan^2 x)=4+4sin^2 x$ (as $sec^2 x-\\tan^2 x = 1$)
\n$\\frac{(1+ sin^2 x)}{cos^2 x} = 4(1+sin^2 x) $
\n$cos^2 x = \\frac{1}{4}$
\n$ cos x = \\frac{1}{2} or \\frac{-1}{2}$
\nHere, $cosx$ cannot be negative as (0<x<90\u00b0)<\/p>\n

Hence, $ cos x = \\frac{1}{2}$ and $ x = 60$\u00b0<\/p>\n

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

$\\sqrt{\\frac{1+cosx}{1-cosx}}$
\nMultiplying the numerator and denominator by 1+cosx, we get
\n$\\sqrt{\\frac{(1+cosx)^2}{(1-cosx)(1+cosx)}}$
\n=> $\\sqrt{\\frac{(1+cosx)^2}{(1- cos^2x}}$
\n=> $\\frac{1+cosx}{sinx}$ = cosecx + cotx$<\/p>\n

 <\/p>\n

GK Questions And Answers PDF<\/a><\/p>\n

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

$sinx = 1 – cos x$
\n=> $sinx + cos x = 1 $
\nSquaring both sides we get,
\n$sin^2x + cos^2x + 2 sinx cos x = 1$
\n=> $1 + sin2x = 1$
\n=> sin 2x = 0 or x=0
\n=> cos 2x = 1<\/p>\n

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

$\\tan\\theta + \\cot\\theta = 2$
\nCubing both sides we get
\n$\\tan^3\\theta + \\cot^3\\theta + 3\\tan\\theta*\\cot\\theta(\\tan\\theta+\\cot\\theta) = 8$
\n=> $\\tan^3\\theta + \\cot^3\\theta + 3*2 = 8$
\n=> $\\tan^3\\theta + \\cot^3\\theta = 2$
\n<\/span>
\n<\/span><\/p>\n

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

$\\sin x+\\frac{1}{\\sin x}=\\frac{5}{2}$
\n=>$2\\sin^2 x-5\\sin x+2=0$
\n=>$(\\sin x-1\/2)(\\sin x-2)=0$
\n=>$\\sin x = 1\/2$
\n=>$x=30$ degrees
\n=>$\\tan x = \\frac{1}{\\sqrt3}$
\n=>$\\cot x = \\sqrt3$
\nThe given expression would be,
\n$\\tan {x}+\\cot {x}$
\n$=\\frac{1}{\\sqrt3}+\\sqrt3$
\n$=\\frac{4}{\\sqrt3}$<\/p>\n

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

$\\cos x+\\frac{1}{\\cos x}=2$
\n=>$\\cos^2 x+1=2\\cos x$
\n=>$(\\cos x-1)^2=0$
\n=>$\\cos x = 1$
\n=>$\\sin x = 0$
\nThe given expression is,
\n$\\sin ^5 x+\\sec ^{-5} x$
\n$= \\sin ^5 x+\\frac{1}{\\cos^5 x}$
\n$=0+1 = 1$<\/p>\n

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

$\\tan x+\\frac{1}{\\tan x}=2$
\n=>$\\tan^2 x+1=2\\tan x$
\n=>$(\\tan x-1)^2=0$
\n=>$\\tan x = 1$
\n=>$\\sin x = \\frac{1}{\\sqrt{2}}$
\n=>$\\cos x = \\frac{1}{\\sqrt{2}}$
\nThe given expression would be,
\n$ \\sin ^5 x+\\cos ^{5} x$
\n$ \\sin ^5 x+\\cos^5 {x}$
\n$=\\frac{1}{4\\sqrt{2}}+\\frac{1}{4\\sqrt{2}} = \\frac{1}{2\\sqrt{2}}$<\/p>\n

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

for asin \u03b8 + bcos \u03b8 + c,
\nmaximum value = $c+\\sqrt{a^{2}+b^{2}}$
\nminimum value = $c-\\sqrt{a^{2}+b^{2}}$
\nfor sin \u03b8 + cos \u03b8 , a = 1, b = 1, c = 0
\nmaximum value = $c+\\sqrt{a^{2}+b^{2}}=0+\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$
\nso the answer is option B.<\/p>\n

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

$sin \\Theta + cosec \\Theta = 5$
\n==> $(sin \\Theta + cosec \\Theta)^3 = 5^3$
\n==>$ sin^3 \\Theta + cosec ^3 \\Theta + 3*sin \\Theta * cosec \\Theta (sin \\Theta + cosec \\Theta)$ = 125
\n==> $sin^3 \\Theta + cosec ^3 \\Theta + 3*5 = 125$
\n==> $sin^3 \\Theta + cosec ^3 \\Theta $ = 110.<\/p>\n

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

Substituting values of angles, we get,
\n3\/4 + 3\/4+ 1 + 4 + 1 = 7.5. Option D is the right answer.<\/p>\n

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

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

$cos x + cos^2 x = 1$
\n=> $cos x = 1 – cos^2 x$
\n=> $cos x = sin^2 x$
\n$\\therefore$ $sin^{8} x + 2 sin^{6} x + sin^{4} x$
\n= $(sin^4 x + sin^2 x)^2$
\n<\/span>= $((cos x)^2 + sin^2 x)^2$
\n<\/span>= $(cos^2 x + sin^2 x)^2 = 1$<\/span><\/p>\n

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

$x= p$ $cosec \u03b8$
\n=> $cosec\\theta$ = $\\frac{x}{p}$
\nAlso, $y= q$ $cot \u03b8$
\n=> $cot\\theta$ = $\\frac{y}{q}$
\n<\/span>$\\because$ $cosec^2\\theta-cot^2\\theta$ = 1
\n<\/span>=> $\\frac{x^2}{p^2}$ – $\\frac{y^2}{q^2}$ = 1<\/span><\/p>\n

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

Expression : tan 4\u00b0 tan 43\u00b0 tan 47 tan 86\u00b0
\n$\\because$ $tan(90-\\theta) = cot\\theta$
\n=> $tan 4^{\\circ} = tan(90^{\\circ}-86^{\\circ}) = cot 86^{\\circ}$
\nSimilarly, $tan 43^{\\circ} = cot 47^{\\circ}$
\n=> $(cot 86^{\\circ} \\times tan 86^{\\circ}) * (tan 47^{\\circ} \\times cot 47^{\\circ})$
\nUsing, $tan\\theta cot\\theta$ = 1
\n=> 1*1 = 1<\/p>\n

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

Expression : $x cos\u03b8 – sin\u03b8 = 1$
\n=> $x = \\frac{1}{cos\\theta} + \\frac{sin\\theta}{cos\\theta}$
\n=> $x = sec\\theta + tan\\theta$ ————–Eqn(1)
\n$\\because$ $sec^2\\theta – tan^2\\theta = 1$
\n=> $(sec\\theta + tan\\theta)(sec\\theta – tan\\theta) = 1$
\n=> $(sec\\theta – tan\\theta) = \\frac{1}{x}$ ————–Eqn(2)
\nAdding eqns(1)&(2)
\n=> $2sec\\theta = x + \\frac{1}{x} = \\frac{x^2 + 1}{x}$
\n=> $sec\\theta = \\frac{x^2 + 1}{2x}$
\nSubtracting eqn(2) from (1)
\n=> $2tan\\theta = x – \\frac{1}{x} = \\frac{x^2 – 1}{x}$
\n=> $tan\\theta = \\frac{x^2 – 1}{2x}$
\nWe know that, $sin\\theta = \\frac{tan\\theta}{sec\\theta}$
\n=> $sin\\theta = \\frac{x^2 – 1}{2x} * \\frac{2x}{x^2 + 1}
\n=> $sin\\theta = \\frac{x^2 – 1}{x^2 + 1}$
\nTo find : $x^2 – (1 + x^2) sin\u03b8 $
\n= $x^2 – (1 + x^2) * \\frac{x^2 – 1}{x^2 + 1}$
\n= $x^2 – (x^2 – 1)$
\n= 1<\/p>\n

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

Expression : $sin\\theta + sin^2\\theta = 1$
\n=> $sin\\theta = 1 – sin^2\\theta$
\n=> $sin\\theta = cos^2\\theta$
\nTo find : $cos^2\\theta + cos^4\\theta$
\n= $cos^2\\theta + (cos^2\\theta)^2$
\n= $cos^2\\theta + sin^2\\theta$
\n= 1<\/p>\n

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

Expression : tan1\u00b0tan2\u00b0tan3\u00b0 \u2026\u2026\u2026\u2026\u2026tan88\u00b0tan89\u00b0
\n$\\because$ $tan(90^{\\circ}-\\theta) = cot\\theta$
\n=> tan 89\u00b0 = tan(90\u00b0-1) = cot 1\u00b0
\nSimilarly, tan 88\u00b0 = cot 2\u00b0 and so on till tan 46\u00b0 = cot 44\u00b0
\n=> (tan1\u00b0tan2\u00b0tan3\u00b0…….tan45\u00b0……cot3\u00b0cot2\u00b0cot1\u00b0)
\nUsing, $tan\\theta cot\\theta$ = 1 and $tan45^{\\circ}$ = 1
\n=> 1*1 = 1<\/p>\n

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

Expression : $sin A cos A (tan A – cot A)$
\n= $sin A cos A (\\frac{sin A}{cos A} – \\frac{cos A}{sin A})$
\n= $sin A cos A (\\frac{sin^2 A – cos^2 A}{sin A cos A})$
\n= $sin^2 A – cos^2 A$
\n= $sin^2 A – (1 – sin^2 A)$
\n= $2sin^2 A – 1$<\/p>\n

 <\/p>\n

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

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

Trigonometry Questions for SSC CHSL Exam PDF: SSC CHSL Trignometry Questions download PDF based on previous year question paper of SSC CHSL exam. 25 Very important Trignometry questions for SSC CHSL 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":21,"featured_media":25823,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,378],"tags":[],"class_list":{"0":"post-25818","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc-chsl"},"better_featured_image":{"id":25823,"alt_text":"Trigonometry Questions for SSC CHSL Exam PDF","caption":"Trigonometry Questions for SSC CHSL Exam PDF","description":"Trigonometry Questions for SSC CHSL Exam 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