{"id":42770,"date":"2020-09-18T19:57:07","date_gmt":"2020-09-18T14:27:07","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42770"},"modified":"2020-09-18T19:57:07","modified_gmt":"2020-09-18T14:27:07","slug":"trigonometry-questions-for-ssc-cgl-tier-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/trigonometry-questions-for-ssc-cgl-tier-2-pdf\/","title":{"rendered":"Trigonometry Questions for SSC CGL Tier 2 PDF"},"content":{"rendered":"

Trigonometry Questions for SSC CGL Tier 2 PDF<\/strong><\/span><\/h2>\n

Download SSC CGL Tier 2 Trigonometry Questions PDF. Top 15 SSC CGL Tier 2 Trigonometry questions based on asked questions in previous exam papers very important for the SSC exam.<\/p>\n

Download Trigonometry Questions for SSC CGL Tier 2 PDF<\/a><\/p>\n

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Question 1:\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 2:\u00a0<\/b>If a cos \u03b8 + b sin \u03b8 = p and a sin \u03b8 – b cos \u03b8 = q, then the relation between a, b, p and q is<\/p>\n

a)\u00a0$a^{2}- b^{2} = p^{2} – q^{2}$<\/p>\n

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

c)\u00a0$a+b=p+q$<\/p>\n

d)\u00a0$a- b=p-q$<\/p>\n

Question 3:\u00a0<\/b>If $2\\sin\\theta+\\cos\\theta-\\frac{7}{3}$ then the value of $(\\tan^{2}\\theta-\\sec^{2}\\theta)$ is <\/span><\/p>\n

a)\u00a00<\/p>\n

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

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

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

Question 4:\u00a0<\/b>If tan \u03b8 + cot \u03b8 = 2, then the value of $tan^n \\theta + cot^{n}\\theta$(0\u00b0 < \u03b8 < 90\u00b0, n is an integer) is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a02n+1<\/p>\n

c)\u00a02n<\/p>\n

d)\u00a00<\/p>\n

Question 5:\u00a0<\/b>If sin \u03b8 + cosec \u03b8 = 2 then the value of $sin^{5}\\theta+cosec^{5}\\theta$ is<\/p>\n

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

b)\u00a01<\/p>\n

c)\u00a00<\/p>\n

d)\u00a02<\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

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Question 6:\u00a0<\/b>Find the value of $1 – 2 sin^{2} \u03b8 + sin^{4} \u03b8.$<\/p>\n

a)\u00a0$sin^{4} \u03b8$<\/p>\n

b)\u00a0$cos^{4} \u03b8$<\/p>\n

c)\u00a0$cosec^{4} \u03b8$<\/p>\n

d)\u00a0$sec^{4} \u03b8$<\/p>\n

Question 7:\u00a0<\/b>If sin (x + y) = cos [3(x + y)], then the value of tan[2(x + y)] is :<\/p>\n

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

b)\u00a01<\/p>\n

c)\u00a00<\/p>\n

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

Question 8:\u00a0<\/b>The height of a tower is h and the angle of elevation of the top of the tower is a. On moving a distance h\/2 towards, the tower, the angle of elevation becomes 0. The value of cot\u03b1 – cot \u03b2 is<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

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

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

Question 9:\u00a0<\/b>If A and B are positive acute angles such that sin (A \u2014 B) =1\/2 and cos (A+ B) = 1\/2 , then A and B are given by<\/p>\n

a)\u00a0A = 45\u00b0, B = 15\u00b0<\/p>\n

b)\u00a0A = 15\u00b0, B = 45\u00b0<\/p>\n

c)\u00a0A = 30\u00b0, B = 30\u00b0<\/p>\n

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

Question 10:\u00a0<\/b>If 2secA – (1 + sinA)\/cosA = x, then the value of x is<\/p>\n

a)\u00a0cosecA\/(1+sinA)<\/p>\n

b)\u00a0cosA\/(1+sinA)<\/p>\n

c)\u00a0cosA(1+sinA)<\/p>\n

d)\u00a0cosecA(1+sinA)<\/p>\n

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

Free SSC Preparation (Videos Youtube)<\/a><\/p>\n

Question 11:\u00a0<\/b>If 1\/(cosecA \u00ad+ cotA) = x, then the value of x is<\/p>\n

a)\u00a0cosecA + cotA<\/p>\n

b)\u00a0cosecA \u00ad- cotA<\/p>\n

c)\u00a0$cosec^{2}$ A + cot2A<\/p>\n

d)\u00a0\u221a[$cosec^{2}$ A + cot2A]<\/p>\n

Question 12:\u00a0<\/b>cos3A is equal to<\/p>\n

a)\u00a0$cos^{3}A – 3sin^{2}cosA$<\/p>\n

b)\u00a0$cos^{3}A + 4sin^{2}cosA$<\/p>\n

c)\u00a0$cos^{3}A + 3sin^{2}cosA$<\/p>\n

d)\u00a0$cos^{3}A – 4sin^{2}cosA$<\/p>\n

Question 13:\u00a0<\/b>What is the value of tan $(-\\frac{5\\pi}{6})$<\/p>\n

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

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

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

d)\u00a0$-{\\sqrt{3}}$<\/p>\n

Question 14:\u00a0<\/b>If cos 135\u00b0 = x, then the value of x is<\/p>\n

a)\u00a0-1\/\u221a2<\/p>\n

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

c)\u00a0-1\/2<\/p>\n

d)\u00a02<\/p>\n

Question 15:\u00a0<\/b>2cos[(C+D)\/2].cos[(C-D)\/2] is equal to<\/p>\n

a)\u00a0cosC – cosD<\/p>\n

b)\u00a0sinC + sinD<\/p>\n

c)\u00a0cosC + cosD<\/p>\n

d)\u00a0sinC – sinD<\/p>\n

Download SSC CGL General Science Notes PDF<\/a><\/p>\n

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

1)\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

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

Expression 1 : a cos \u03b8 + b sin \u03b8 = p<\/p>\n

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

=> $a^2 cos^2 \\theta + b^2 sin^2 \\theta + 2ab sin\\theta cos\\theta = p^2$ ——–Eqn(1)<\/p>\n

Expression 2 : a sin \u03b8 – b cos \u03b8 = q<\/p>\n

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

=> $a^2 sin^2 \\theta + b^2 cos^ \\theta – 2ab sin\\theta cos\\theta = q^2$ ———-Eqn(2)<\/span><\/p>\n

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

=> $a^2 (sin^2 \\theta+cos^2 \\theta) + b^2 (sin^2 \\theta + cos^2 \\theta) = p^2 + q^2$<\/span><\/p>\n

=> $a^{2} + b^{2} = p^{2}+ q^{2}$<\/span><\/p>\n

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

tan2 <\/sup>\u03b8 – sec2<\/sup>\u03b8<\/p>\n

\u21d2 – (sec2<\/sup>\u03b8 – tan2 <\/sup>\u03b8)<\/p>\n

\u21d2 -1 because sec2<\/sup>\u03b8 – tan2 <\/sup>\u03b8 = 1<\/p>\n

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

Given tan \u03b8 + cot \u03b8 = 2
\nThen $ (tan \u03b8 + cot \u03b8)^2 = 4$
\n$ (tan ^2\u03b8 + cot ^2\u03b8 + 2tan \u03b8 cot \u03b8) = 4$
\n$ (tan ^2\u03b8 + cot ^2\u03b8 ) = 2$
\nOption A is the correct answer.<\/span>
\n<\/span><\/p>\n

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

Since $cosec\u03b8 = \\frac{1}{sin\u03b8}$
\n<\/span>$sin \u03b8 + cosec \u03b8 = 2 $ becomes
\n$sin \u03b8 + \\frac{1}{sin\u03b8} =2$
\n$sin^2\u03b8 – 2sin\u03b8 +1 =0 $
\nwhich is $(sin\u03b8 – 1)^2 = 0$
\n$sin \u03b8 =1$
\n$sin^{5}\\theta+cosec^{5}\\theta = 1 + 1 = 2$
\nHence Option D is th correct answer.
\n<\/span><\/span><\/span><\/span><\/p>\n

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

Here,<\/p>\n

$1 – 2 sin^{2} \u03b8 + sin^{4} \u03b8.$ = $1^2 + (sin^2 \\theta)^2 – 2 \\times 1 \\times sin^2 \\theta$<\/p>\n

it is similar to $(a-b)^2$ = $a^2 + b^2 – 2ab$<\/p>\n

So,<\/p>\n

$1^2 + (sin^2 \\theta)^2 – 2 \\times 1 \\times sin^2 \\theta$ = $(sin^2 \\theta – 1)^2$……(1)
\n<\/span><\/p>\n

Now $sin^2 \\theta + cos^2 \\theta$ = 1…..(2)<\/span><\/p>\n

From equation 1 and 2 <\/span><\/p>\n

$(sin^2 \\theta – 1)^2$ = $(sin^2 \\theta – sin^2 \\theta – cos^2 \\theta)^2$
\n<\/span><\/span><\/p>\n

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

= $cos^4 \\theta$<\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

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

Given,<\/p>\n

sin (x + y) = cos [3(x + y)]
\nUsing: cos \u03b8 = sin (90\u00b0 – \u03b8)
\nsin (x + y) = sin [90\u00b0 – 3(x + y)]
\nsin [90\u00b0 – 3(x + y)] – sin (x + y) = 0
\nsin<\/span>C<\/span>\u2212<\/span>sin<\/span>D <\/span>=<\/span>2<\/span>sin[(<\/span>C<\/span>\u2212<\/span>D)\/<\/span><\/span><\/span>2] <\/span><\/span>cos<\/span>[(<\/span>C+<\/span>D)\/<\/span><\/span><\/span>2]<\/span><\/span><\/span><\/span><\/span><\/span>
\n=2sin$\\frac{(90-3(x+y)-(x+y))}{2} cos \\frac{90-3(x+y)+(x+y)}{2}$=0<\/span><\/span><\/span>
\n=2sin(45-2(x+y)) cos (45-(x+y))=0<\/span><\/span><\/span><\/span><\/span><\/span>
\n\u2234 sin 45\u00b0 – 2(x + y)} = 0
\n45\u00b0 – 2(x + y) = 0
\n2(x + y) = 45\u00b0
\nOR
\nCos{45\u00b0- (x + y)} = 0
\n45\u00b0- (x + y)} = 90\u00b0
\nx + y = – 45\u00b0
\n2(x + y) = – 90\u00b0
\nPutting 2(x + y) = 45\u00b0
\ntan 2(x + y) = tan 45\u00b0 = 1
\nAgain, Putting 2(x + y) = – 90\u00b0, we will not get any answer among given options
\nOption B is the correct answer.<\/p>\n

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

<\/p>\n

Here, $\\angle$ACB = $\\alpha$ and $\\angle$ADB = $\\beta$<\/p>\n

AB = tower = $h$ metre<\/p>\n

and CD = $\\frac{h}{2}$ metre<\/p>\n

From $\\triangle$ABC<\/p>\n

=> $tan \\alpha = \\frac{AB}{BC} = \\frac{h}{BC}$<\/p>\n

=> $BC = h cot \\alpha$ ———-Eqn(1)<\/p>\n

From $\\triangle$ABD<\/p>\n

=> $tan \\beta = \\frac{AB}{BD} = \\frac{h}{BC – CD}$<\/p>\n

=> $tan \\beta = \\frac{h}{h cot \\alpha – \\frac{h}{2}}$<\/p>\n

=> $h cot \\alpha – \\frac{h}{2} = h cot \\beta$<\/p>\n

=> $h (cot \\alpha – cot \\beta) = \\frac{h}{2}$<\/p>\n

=> $cot \\alpha – cot \\beta = \\frac{1}{2}$<\/p>\n

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

$sin (A – B) = \\frac{1}{2} = sin 30^{\\circ}$<\/p>\n

=> $A – B = 30^{\\circ}$ ———-Eqn(1)<\/p>\n

Again, $cos (A + B) = \\frac{1}{2} = cos 60^{\\circ}$<\/p>\n

=> $A + B = 60^{\\circ}$ ———Eqn(2)<\/p>\n

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

$2A = 90^{\\circ}$<\/p>\n

=> $A = 45^{\\circ}$ and $B = 15^{\\circ}$<\/p>\n

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

Expression\u00a0:\u00a02secA – (1 + sinA)\/cosA = x<\/p>\n

= $\\frac{2}{sec A} – \\frac{(1 + sin A)}{cos A}$<\/p>\n

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

Multiplying both numerator and denominator by $(1 + sin A)$<\/p>\n

= $\\frac{1 – sin A}{cos A} \\times \\frac{(1 + sin A)}{(1 + sin A)}$<\/p>\n

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

= $\\frac{cos A}{1 + sin A}$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0: $\\frac{1}{cosec A + cot A}$<\/p>\n

= $\\frac{1}{\\frac{1}{sin A} + \\frac{cos A}{sin A}}$<\/p>\n

= $\\frac{1}{\\frac{1 + cos A}{sin A}} = \\frac{sin A}{1 + cos A}$<\/p>\n

Multiplying both numerator and denominator by $(1 – cos A)$<\/p>\n

= $\\frac{sin A}{1 + cos A} \\times \\frac{(1 – cos A)}{(1 – cos A)}$<\/p>\n

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

= $\\frac{1 – cos A}{sin A} = \\frac{1}{sin A} – \\frac{cos A}{sin A}$<\/p>\n

= $cosec A – cot A$<\/p>\n

=> Ans – (B)<\/p>\n

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

Using triple angle formula, we know that\u00a0: $cos(3A) = 4cos^3A – 3cosA$<\/p>\n

= $cos^3A + (3cos^3A – 3cosA)$<\/p>\n

= $cos^3A + 3cosA(cos^2A – 1)$<\/p>\n

= $cos^3A – 3cosA(1 – cos^2A)$<\/p>\n

= $cos^3A – 3sin^2AcosA$<\/p>\n

=> Ans – (A)<\/p>\n

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

Expression\u00a0:\u00a0tan $(-\\frac{5\\pi}{6})$<\/p>\n

= $- tan(\\frac{5 \\pi}{6})$<\/p>\n

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

= $tan(\\frac{\\pi}{6}) = \\frac{1}{\\sqrt{3}}$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0: cos 135\u00b0 = x<\/p>\n

= $cos (180 – 45) = -cos (45$\u00b0$)$<\/p>\n

= $\\frac{-1}{\\sqrt{2}}$<\/p>\n

=> Ans – (A)<\/p>\n

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

Expression\u00a0:\u00a02cos[(C+D)\/2]cos[(C-D)\/2]<\/p>\n

Using the formula\u00a0: $cos x . cos y = \\frac{1}{2} [cos (x + y) + cos (x – y)]$ ————–(i)<\/p>\n

Substituting $(x + y) = C$ and $(x – y) = D$<\/p>\n

=> $x = \\frac{C + D}{2}$ and $y = \\frac{C – D}{2}$ in equation (i),<\/p>\n

=> $cos (\\frac{C + D}{2}) . cos (\\frac{C – D}{2}) = \\frac{1}{2} [cos C + cos D]$<\/p>\n

=> $2 . cos (\\frac{C + D}{2}) . cos (\\frac{C – D}{2}) = cos C + cos D$<\/p>\n

=> Ans – (C)<\/p>\n

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We hope this Trogonometry Questions pdf for SSC CGL Tier 2 exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Trigonometry Questions for SSC CGL Tier 2 PDF Download SSC CGL Tier 2 Trigonometry Questions PDF. Top 15 SSC CGL Tier 2 Trigonometry questions based on asked questions in previous exam papers very important for the SSC exam. Take SSC CGL Tier-2 Mock Tests Download SSC CGL Tier-2 Previous Papers PDF Question 1:\u00a0If 0\u00b0 \u2264 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