GET 20 SSC GD MOCK FOR JUST RS. 117<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Download SSC GD Important Questions PDF<\/a><\/p>\n 1500<\/strong>+ Must Solve Questions for SSC Exams<\/a> (Question bank)<\/p>\n Question 1:\u00a0<\/b>If $\\sec \\theta +\\tan \\theta=2$, then the value of $\\sin \\theta$ is:<\/p>\n a)\u00a0$\\frac{4}{5}$<\/p>\n b)\u00a0$\\frac{\\sqrt{3}}{5}$<\/p>\n c)\u00a0$\\frac{2}{5}$<\/p>\n d)\u00a0$\\frac{3}{5}$<\/p>\n Question 2:\u00a0<\/b>If $\\alpha + \\theta$ = $\\frac{7\\pi}{12}$ and $\\tan \\theta = \\sqrt{3}$, then the value of $\\tan \\theta$ 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 3:\u00a0<\/b>If $\\cos\\theta + \\sec\\theta = \\sqrt{3}$, then the value of $(\\cos^{3}\\theta + \\sec^{3}\\theta)$ is:<\/p>\n a)\u00a01<\/p>\n b)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n c)\u00a00<\/p>\n d)\u00a0${\\sqrt{2}}$<\/p>\n Question 4:\u00a0<\/b>The value of the following is: $\\frac{\\sin\\theta\\ \\cos ec\\theta\\ \\tan\\theta\\ \\cot\\theta}{\\sin^{2}\\theta+\\cos^{2}\\theta}$<\/p>\n a)\u00a01<\/p>\n b)\u00a0$\\tan\\theta$<\/p>\n c)\u00a00<\/p>\n d)\u00a02<\/p>\n Question 5:\u00a0<\/b>In a right angled triangle ABC and right angled at B, AB=BC, what is the value of sinA$\\star$cosC ?<\/p>\n a)\u00a0$\\frac{1}{4}$<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a0$\\frac{1}{6}$<\/p>\n d)\u00a0$\\frac{1}{5}$<\/p>\n Download SSC GD FREE PREVIOUS PAPERS<\/a><\/p>\n LATEST FREE SSC GD MOCK 2019<\/a><\/p>\n Question 6:\u00a0<\/b>If $\\sin\\theta + \\frac{1}{\\sin\\theta} =\\frac{5}{2}$ and $\\theta\\in Q1$ then what is the value of $\\tan\\theta ?$<\/p>\n a)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n b)\u00a0$1$<\/p>\n c)\u00a0$\\sqrt{3}$<\/p>\n d)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n Question 7:\u00a0<\/b>The elevation of the top of a tower from a point on the ground is $\\ 45^\\circ$. On travelling 60 m from the point towards the tower, the alevation of the top becomes $\\ 60^\\circ$. The height of the tower, in metres, is<\/p>\n a)\u00a030<\/p>\n b)\u00a030(3- $\\ \\sqrt{3}$)<\/p>\n c)\u00a030(3 +$\\ \\sqrt{3}$)<\/p>\n d)\u00a030$\\ \\sqrt{3}$<\/p>\n Question 8:\u00a0<\/b>The expression $\\sqrt{\\frac{1+sin\\ \\theta}{1-sin\\ \\theta}}+\\sqrt{\\frac{1-sin\\ \\theta}{1+sin\\ \\theta}}$ is equal to<\/p>\n a)\u00a0$2\\ sec\\ \\theta$<\/p>\n b)\u00a0$2\\ tan\\ \\theta$<\/p>\n c)\u00a0$\\frac{2\\ (1-sin\\ \\theta)}{cos\\ \\theta}$<\/p>\n d)\u00a0$2\\ sin\\ \\theta$<\/p>\n Question 9:\u00a0<\/b>If $13sinA = 12$ then find out the value of $\\frac{3sinA + 4cosA}{3sinA – 4cosA}$? Assume that ($0 \\leq A \\leq 90$)<\/p>\n a)\u00a0$\\frac{7}{2}$<\/p>\n b)\u00a0$\\frac{7}{4}$<\/p>\n c)\u00a0$\\frac{7}{16}$<\/p>\n d)\u00a0$\\frac{7}{8}$<\/p>\n Question 10:\u00a0<\/b>What is the minimum and maximum value that the expression $12Sin\\theta – 35cos\\theta$ can take?<\/p>\n a)\u00a0$-23, 47$<\/p>\n b)\u00a0$-47, 23$<\/p>\n c)\u00a0$-47, 47$<\/p>\n d)\u00a0$-37, 37$<\/p>\n DOWNLOAD APP TO ACESS DIRECTLY ON MOBILE<\/a><\/p>\n Daily Free SSC Practice Set<\/a><\/p>\n Question 11:\u00a0<\/b>If $cos\\theta_{1} = \\frac{35}{37}$, $sin\\theta_{2} = \\frac{3}{5}$,then find out $\\tan(\\theta_{2} – \\theta_{1})$ ? (Assume that $0 \\leq \\theta_{1},\\theta_{2} \\leq 90$)<\/p>\n a)\u00a0$\\frac{49}{71}$<\/p>\n b)\u00a0$\\frac{57}{76}$<\/p>\n c)\u00a0$\\frac{75}{176}$<\/p>\n d)\u00a0$\\frac{114}{352}$<\/p>\n Question 12:\u00a0<\/b>If $tan\\theta = \\frac{11}{60}$ then find out $\\frac{cos\\theta – sin\\theta}{cos\\theta + sin\\theta}$ ?<\/p>\n a)\u00a0$\\frac{49}{71}$<\/p>\n b)\u00a0$\\frac{55}{65}$<\/p>\n c)\u00a0$\\frac{65}{55}$<\/p>\n d)\u00a0$\\frac{71}{64}$<\/p>\n Question 13:\u00a0<\/b>If $tan \u03b8 + cot \u03b8 = x$, then what is the value of $tan^4\u03b8 + cot^4\u03b8$?<\/p>\n a)\u00a0$x^2(x^2-4)+6$<\/p>\n b)\u00a0$x^2(x^2-4)+2$<\/p>\n c)\u00a0$x^2(x^2-6)+2$<\/p>\n d)\u00a0$x^2(x^2-2)+2$<\/p>\n Question 14:\u00a0<\/b>If $sin^2\\theta – cos^2\\theta = \\frac{16}{25}$, what is the value of $cos^4\\theta – sin^4\\theta$?<\/p>\n a)\u00a0$-\\frac{4}{25}$<\/p>\n b)\u00a0$-\\frac{16}{5}$<\/p>\n c)\u00a0$-\\frac{16}{25}$<\/p>\n d)\u00a0$-\\frac{4}{5}$<\/p>\n Question 15:\u00a0<\/b>If $cosec^2\\theta + cot^2\\theta = \\frac{11}{5}$, what is the value of $cosec^4\\theta – cot^4\\theta$?<\/p>\n a)\u00a0$\\frac{11}{5}$<\/p>\n b)\u00a0$\\frac{121}{25}$<\/p>\n c)\u00a0$\\frac{5}{11}$<\/p>\n d)\u00a0$\\frac{11}{25}$<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n Download SSC GD Constable Syllabus PDF<\/a><\/p>\n Question 16:\u00a0<\/b>If $tan^2\\theta + sec^2\\theta = \\frac{9}{8}$, what is the value of $ tan^4\\theta – sec^4\\theta $?<\/p>\n a)\u00a0$-\\frac{81}{64}$<\/p>\n b)\u00a0$-\\frac{9}{8}$<\/p>\n c)\u00a0$-\\frac{81}{8}$<\/p>\n d)\u00a0$-\\frac{9}{64}$<\/p>\n Question 17:\u00a0<\/b>If $cos \u03b8 + sec \u03b8 = a$, then what is the value of $cos^6\u03b8 + sec^6\u03b8$?<\/p>\n a)\u00a0$a^6 + 6a^4 + 6a^2 + 2$<\/p>\n b)\u00a0$a^6 – 6a^4 + 18a^2 – 14$<\/p>\n c)\u00a0$a^6 – 6a^4 + 18a^2 + 2$<\/p>\n d)\u00a0$a^6 – 6a^4 + 6a^2 – 2$<\/p>\n Question 18:\u00a0<\/b>Shadow of a tower is 60 m long and the towers\u2019s elevation angle is 60\u00b0 then what is the height of the tower?<\/p>\n a)\u00a0120 m<\/p>\n b)\u00a0$60\\sqrt{3}$ m<\/p>\n c)\u00a0$\\frac{60}{\\sqrt{3}}$ m<\/p>\n d)\u00a0None of these<\/p>\n Question 19:\u00a0<\/b>If $tanA-cotA =0$ then find out the value of ($tan^4A+cot^4A$) ?<\/p>\n a)\u00a02<\/p>\n b)\u00a04<\/p>\n c)\u00a00<\/p>\n d)\u00a0None of these<\/p>\n Question 20:\u00a0<\/b>If $5tanA = 12$ then find out the value of $\\frac{12sinA + 5cosA}{12sinA – 5cosA}$?<\/p>\n a)\u00a0$\\frac{129}{179}$<\/p>\n b)\u00a0$\\frac{179}{129}$<\/p>\n c)\u00a0$\\frac{119}{169}$<\/p>\n d)\u00a0$\\frac{169}{119}$<\/p>\n Question 21:\u00a0<\/b>It is given that sinA = $\\frac{-\\sqrt{3}}{2}$, cosB = $\\frac{\\sqrt{3}}{2}$ then what is the value of (A+B) ? a)\u00a0600\u00b0<\/p>\n b)\u00a0630\u00b0<\/p>\n c)\u00a0570\u00b0<\/p>\n d)\u00a0More than one of the above<\/p>\n Question 22:\u00a0<\/b>It is given that cosA = -$\\frac{1}{\\sqrt{2}}$, cosB = -$\\frac{1}{2}$, then what is the value of (A+B) ? a)\u00a0465\u00b0<\/p>\n b)\u00a0435\u00b0<\/p>\n c)\u00a0150\u00b0<\/p>\n d)\u00a0More than one of the above<\/p>\n Question 23:\u00a0<\/b>It is given that sinA = $\\frac{\\sqrt{3}}{2}$, cosB = $\\frac{1}{2}$ then what is the value of (A+B) ? a)\u00a0120\u00b0<\/p>\n b)\u00a090\u00b0<\/p>\n c)\u00a0150\u00b0<\/p>\n d)\u00a0More than one of the above<\/p>\n Question 24:\u00a0<\/b>If $\\cos\\theta_{1} = \\frac{\\sqrt{3}}{2}$, $\\cos\\theta_{2} = \\frac{24}{25}$ then find out the value of $\\tan(\\theta_{1}-\\theta_{2})$ ? (Assume that $0 \\leq \\theta_{1},\\theta_{2} \\leq 90$)<\/p>\n a)\u00a0$\\frac{7 + 24\\sqrt{3}}{7 – 24\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{7 – 24\\sqrt{3}}{7 + 24\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{24 – 7\\sqrt{3}}{24\\sqrt{3} + 7}$<\/p>\n d)\u00a0$\\frac{24 +7\\sqrt{3}}{24\\sqrt{3} – 7}$<\/p>\n Question 25:\u00a0<\/b>If $\\cos\\theta_{1} = \\frac{1}{\\sqrt{2}}$, $\\cos\\theta_{2} = \\frac{40}{41}$ then find out the value of $\\cos(\\theta_{1}-\\theta_{2})$ ? (Assume that $0 \\leq \\theta_{1},\\theta_{2} \\leq 90$)<\/p>\n a)\u00a0$\\frac{49\\sqrt{2}}{82}$<\/p>\n b)\u00a0$\\frac{49}{82\\sqrt{2}}$<\/p>\n c)\u00a0$\\frac{41\\sqrt{2}}{98}$<\/p>\n d)\u00a0$\\frac{41\\sqrt{2}}{49}$<\/p>\n Question 26:\u00a0<\/b>If $\\sin\\theta_{1} = \\frac{3}{5}$, $\\cos\\theta_{2} = \\frac{7}{25}$ then find out the value of $\\sin(\\theta_{1}+\\theta_{2})$ ? (Assume that $0 \\leq \\theta_{1},\\theta_{2} \\leq 90$)<\/p>\n a)\u00a0$\\frac{99}{125}$<\/p>\n b)\u00a0$\\frac{117}{125}$<\/p>\n c)\u00a0$\\frac{96}{125}$<\/p>\n d)\u00a0$\\frac{107}{125}$<\/p>\n Question 27:\u00a0<\/b>$\\frac{\\sin x}{1 + \\cos x} + \\frac{1 +\\ cos x}{\\sin x} = 4$ for 0\u00b0 < $x$ < 90\u00b0 a)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n b)\u00a0$\\sqrt{3}$<\/p>\n c)\u00a01<\/p>\n d)\u00a00<\/p>\n Question 28:\u00a0<\/b>If $\\cos 4x = \\sin x$ where, $0 < x < 90$, what is the value of $\\sin 5x$?<\/p>\n a)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a00.5<\/p>\n d)\u00a00.33<\/p>\n Question 29:\u00a0<\/b>$2a + a^2\\tan 2x = \\tan 2x$ a)\u00a0$a^2$<\/p>\n b)\u00a0$1 + a^2$<\/p>\n c)\u00a0$a^2 – 1$<\/p>\n d)\u00a0$2a^2 + 1$<\/p>\n Question 30:\u00a0<\/b>If $\\cos\\theta = \\frac{3}{5}$ then find out the value of $\\frac{\\tan\\theta – \\cot\\theta}{\\tan\\theta + \\cot\\theta }$ ? (Assume that $0 \\leq \\theta \\leq 90$)<\/p>\n a)\u00a0$\\frac{16}{25}$<\/p>\n b)\u00a0$\\frac{18}{25}$<\/p>\n c)\u00a0$\\frac{7}{25}$<\/p>\n d)\u00a0$\\frac{9}{25}$<\/p>\n Question 31:\u00a0<\/b>If $\\frac{\\tan^2\\theta+3}{\\tan^2\\theta-3} = 2$, what is the value of $\\frac{\\sec^2\\theta+\\cot^2\\theta}{\\sec^2\\theta-\\cot^2\\theta}$?(Assume $\\theta$ is in 1st quadrant)<\/p>\n a)\u00a0$\\frac{89}{91}$<\/p>\n b)\u00a0$\\frac{114}{71}$<\/p>\n c)\u00a0$\\frac{6}{5}$<\/p>\n d)\u00a0$\\frac{91}{89}$<\/p>\n Question 32:\u00a0<\/b>If $\\cos\\theta = \\frac{\\sqrt{2+\\sqrt{3}}}{2}$, then what is the value of $cos\\theta +sec\\theta$? (Assume that $0 \\leq \\theta \\leq 90$)<\/p>\n a)\u00a0$\\frac{6 +\\sqrt{3}}{2\\sqrt{2+\\sqrt{3}}}$<\/p>\n b)\u00a0$\\frac{3 +\\sqrt{3}}{\\sqrt{2+\\sqrt{3}}}$<\/p>\n c)\u00a0$\\frac{3 +\\sqrt{6}}{\\sqrt{2+\\sqrt{6}}}$<\/p>\n d)\u00a0$\\frac{6 +\\sqrt{6}}{\\sqrt{3+\\sqrt{6}}}$<\/p>\n Question 33:\u00a0<\/b>If $\\frac{\\sec^2\\theta+1}{\\sec^2\\theta-1} = \\frac{37}{35}$, what is the value of $\\frac{\\tan\\theta+\\sin\\theta}{\\tan\\theta-\\sin\\theta}$ ?(Assume $\\theta$ is in 1st quadrant)<\/p>\n a)\u00a0$\\frac{5}{7}$<\/p>\n b)\u00a0$\\frac{7}{5}$<\/p>\n c)\u00a0$\\frac{7}{6}$<\/p>\n d)\u00a0$\\frac{6}{7}$<\/p>\n Question 34:\u00a0<\/b>From the top of 600 m high tower, the angles of depression of the top and bottom of a flag are observed to be 30\u00b0 and 45\u00b0 respectively. What is the height of the flag?<\/p>\n a)\u00a0$200(\\sqrt{3} – 1)$<\/p>\n b)\u00a0$200(\\sqrt{3} – \\sqrt{2})$<\/p>\n c)\u00a0$200\\sqrt{3}(\\sqrt{3} – 1)$<\/p>\n d)\u00a0None of the above<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n Question 35:\u00a0<\/b>If $cos\\theta – sec\\theta = 7$, what is the value of $cos^2\\theta +sec^2\\theta$?<\/p>\n a)\u00a051<\/p>\n b)\u00a036<\/p>\n c)\u00a049<\/p>\n d)\u00a0None of the above<\/p>\n Question 36:\u00a0<\/b>If $\\sin\\theta + \\cosec\\theta = 2$ then find out the value of $\\cosec^{32}\\theta + \\sin^{32}\\theta$ ?<\/p>\n a)\u00a032<\/p>\n b)\u00a016<\/p>\n c)\u00a064<\/p>\n d)\u00a0None of the above<\/p>\n Question 37:\u00a0<\/b>If $\\sec\\theta = \\frac{15}{9}$ then find out the value of $\\frac{\\cot\\theta + \\cosec\\theta}{\\cot\\theta – \\cosec\\theta }$ ? (Assume that $0 \\leq \\theta \\leq 90$)<\/p>\n a)\u00a03<\/p>\n b)\u00a0-3<\/p>\n c)\u00a0-4<\/p>\n d)\u00a04<\/p>\n Question 38:\u00a0<\/b>If sec \u03b8 = 25\/24, then what is the value of cot \u03b8? (Assume that $0 \\leq \\theta \\leq 90$)<\/p>\n a)\u00a0$\\frac{7}{24}$<\/p>\n b)\u00a0$\\frac{25}{7}$<\/p>\n c)\u00a0$\\frac{24}{7}$<\/p>\n d)\u00a0$\\frac{7}{25}$<\/p>\n Question 39:\u00a0<\/b>If cos \u03b8 + sec \u03b8 = a, then what is the value of $cos^4\u03b8 + sec^4\u03b8$ ?<\/p>\n a)\u00a0$a^2(a^2-4) + 2$<\/p>\n b)\u00a0$a(a-4) + 2$<\/p>\n c)\u00a0$(a^3-3)^2+2$<\/p>\n d)\u00a0$(a^4 – 2a) + 4$<\/p>\n Question 40:\u00a0<\/b>If $x + y = z$ and $tan z = 20$, $m = tan x + tan y$ and $n = tan x * tan y$, what is the value of $\\frac{m}{1 – n}$? (Assume that $0 \\leq x, y, z \\leq 90$)<\/p>\n a)\u00a040<\/p>\n b)\u00a020<\/p>\n c)\u00a010<\/p>\n d)\u00a05<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $\\sec \\theta +\\tan \\theta=2$ ———–(i)<\/p>\n Also, $sec^2\\theta-tan^2\\theta=1$<\/p>\n => $(sec\\theta-tan\\theta)(sec\\theta+tan\\theta)=1$<\/p>\n => $\\sec \\theta -\\tan \\theta=\\frac{1}{2}$ ———-(ii)<\/p>\n Adding equations (i) and (ii), => $2sec\\theta=2+\\frac{1}{2}=\\frac{5}{2}$<\/p>\n => $sec\\theta=\\frac{5}{4}$<\/p>\n => $cos\\theta=\\frac{4}{5}$<\/p>\n $\\therefore$ $sin\\theta=\\sqrt{1-cos^2\\theta}$<\/p>\n = $\\sqrt{1-(\\frac{4}{5})^2}=\\sqrt{1-\\frac{16}{25}}$<\/p>\n = $\\sqrt{\\frac{9}{25}}=\\frac{3}{5}$<\/p>\n => Ans – (D)<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $\\cos\\theta + \\sec\\theta = \\sqrt{3}$ ———-(i)<\/p>\n Cubing both sides, we get :<\/p>\n => $(\\cos\\theta + \\sec\\theta)^3 = (\\sqrt{3})^3$<\/p>\n => $cos^3\\theta+sec^3\\theta+3(cos\\theta)(sec\\theta)(cos\\theta+sec\\theta)=3\\sqrt3$<\/p>\n => $cos^3\\theta+sec^3\\theta+3(cos\\theta\\times sec\\theta)(\\sqrt3)=3\\sqrt3$<\/p>\n $\\because$ $cos\\theta\\times sec\\theta=1$ and using equation (i),<\/p>\n => $cos^3\\theta+sec^3\\theta=3\\sqrt3-3\\sqrt3=0$<\/p>\n => Ans – (C)<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $\\frac{\\sin\\theta\\ \\cos ec\\theta\\ \\tan\\theta\\ \\cot\\theta}{\\sin^{2}\\theta+\\cos^{2}\\theta}$<\/p>\n = $\\frac{(sin\\theta\\times\\frac{1}{sin\\theta})\\times(tan\\theta\\times\\frac{1}{tan\\theta})}{1}$<\/p>\n = $1$<\/p>\n => Ans – (A)<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given is a right angled triangle at B.<\/p>\n Since it is a right angled triangle $AC^{2}=AB^{2}+BC^{2}$.<\/p>\n Given AB=BC. cosC=$\\frac{BC}{AC}$<\/p>\n sinA$\\star$cosC=$\\frac{BC^{2}}{AC^{2}}$<\/p>\n Substituting $AC$=$\\sqrt{2}\\star BC$<\/p>\n sinA$\\star$cosC=$\\frac{1}{2}$<\/p>\n SSC GD FREE STUDY MATERIAL<\/a><\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given $\\sin\\theta +\\frac{1}{\\sin\\theta}$=$\\frac{5}{2}$ 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n From $\\triangle$ ACD,<\/p>\n tan $60^\\circ = \\frac{AD}{CD}$<\/p>\n $\\Rightarrow AD = CD\\sqrt{3}$<\/p>\n From $\\triangle$ ABD,<\/p>\n tan $45^\\circ = \\frac{AD}{BD}$<\/p>\n $\\Rightarrow$ AD = BD<\/p>\n $\\Rightarrow$ AD = BC+CD<\/p>\n $\\Rightarrow$ $AD = 60+\\frac{AD}{\\sqrt{3}}$<\/p>\n $\\Rightarrow$ $AD-\\frac{AD}{\\sqrt{3}} = 60$<\/p>\n $\\Rightarrow \\frac{\\sqrt{3}AD-AD}{\\sqrt{3}} = 60$<\/p>\n $AD(\\sqrt{3}-1) = 60\\sqrt{3}$<\/p>\n $AD = \\frac{60\\sqrt{3}}{\\sqrt{3}-1}$<\/p>\n Rationalising above equation<\/p>\n $AD = \\frac{60\\sqrt{3}}{\\sqrt{3}-1}\\times\\frac{\\sqrt{3}+1}{\\sqrt{3}+1}$<\/p>\n $AD = \\frac{60\\sqrt{3}(\\sqrt{3}+1)}{\\sqrt{3}}$<\/p>\n $\\therefore AD = 30(\\sqrt{3}+3)$<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $\\sqrt{\\frac{1+sin\\ \\theta}{1-sin\\ \\theta}}+\\sqrt{\\frac{1-sin\\ \\theta}{1+sin\\ \\theta}}$<\/p>\n = $\\frac{(\\sqrt{1+sin\\theta})^2+(\\sqrt{1-sin\\theta})^2}{(\\sqrt{1-sin\\theta})(\\sqrt{1+sin\\theta})}$<\/p>\n = $\\frac{(1+sin\\theta)+(1-sin\\theta)}{\\sqrt{1-sin^2\\theta}}$<\/p>\n = $\\frac{2}{cos\\theta}=2sec\\theta$<\/p>\n => Ans – (A)<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that $sinA = \\frac{12}{13}$<\/p>\n $\\Rightarrow$ $\\cos^2A = \\sqrt{1 – \\sin^2A }$<\/p>\n $\\Rightarrow$ $\\cos^2A = \\sqrt{1 – (\\frac{12}{13})^2}$<\/p>\n $\\Rightarrow$ $\\cos^2A = \\frac{25}{169}$<\/p>\n $\\Rightarrow$ $cosA$ = $\\frac{5}{13}$ ($0 \\leq A \\leq 90$)<\/p>\n We have to find out $\\frac{3sinA + 4cosA}{3sinA – 4cosA}$<\/p>\n $\\Rightarrow$ $\\frac{3\\times \\frac{12}{13} + 4\\times \\frac{5}{13}}{3\\times \\frac{12}{13} – 4\\times \\frac{5}{13}}$<\/p>\n $\\Rightarrow$ $\\frac{36+20}{36-20}$<\/p>\n $\\Rightarrow$ $\\frac{7}{2}$<\/p>\n Hence option A is the correct answer.<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We know that the minimum and maximum values of the expression $ASin\\theta + Bcos\\theta$ is given by:<\/p>\n Minimum value =$-\\sqrt{A^2+B^2}$<\/p>\n Maximum value =$\\sqrt{A^2+B^2}$<\/p>\n So in the given question: (A=12, B=-35)<\/p>\n Minimum value =$ -\\sqrt{A^2+B^2} = -\\sqrt{(12)^2+(-35)^2} = -37$<\/p>\n Maximum value =$ \\sqrt{A^2+B^2} = \\sqrt{(12)^2+(-35)^2} = 37$<\/p>\n So, the minimum and maximum value of the expression will be $-37$ and $37$.<\/p>\n Therefore, option D is the right answer.<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n It is given that $cos\\theta_{1} = \\frac{35}{37}$<\/p>\n $\\Rightarrow$ $\\sin^2\\theta_{1} = \\sqrt{1 – \\cos^2\\theta_{1} }$<\/p>\n $\\Rightarrow$ $\\sin^2\\theta_{1} = \\sqrt{1 – (\\frac{35}{37})^2}$<\/p>\n $\\Rightarrow$ $\\sin^2\\theta_{1} = \\frac{144}{1369}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{1} = \\frac{12}{37}$ ($0 \\leq \\theta \\leq 90$)<\/p>\n Also $\\tan\\theta_{1} = \\frac{\\sin\\theta_{1} }{cos\\theta_{1} }$<\/p>\n $\\Rightarrow$ $\\tan\\theta_{1} = \\frac{\\frac{12}{37}}{\\frac{35}{37}}$<\/p>\n $\\Rightarrow$ $\\tan\\theta_{1} = \\frac{12}{35}$ … (1)<\/p>\n Also $\\cos^2\\theta_{2} = \\sqrt{1 – \\sin^2\\theta_{2} }$<\/p>\n $\\Rightarrow$ $\\cos^2\\theta_{2} = \\sqrt{1 – (\\frac{3}{5})^2}$<\/p>\n $\\Rightarrow$ $\\cos\\theta_{2} = \\frac{4}{5}$<\/p>\n Also $\\tan\\theta_{2} = \\frac{\\sin\\theta_{1} }{cos\\theta_{1}} = \\frac{3}{4}$ …(2)<\/p>\n $\\tan(\\theta_{2} – \\theta{1})$ = $\\frac{\\tan\\theta_{2} – \\tan\\theta_{1}}{1 + \\tan\\theta_{2}*\\tan\\theta_{1}}$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{3}{4} – \\frac{12}{35}}{1+\\frac{3}{4}*\\frac{12}{35}}$<\/p>\n $\\Rightarrow$ $\\frac{57}{176}$<\/p>\n Multiplying the numerator and denominator by 2<\/p>\n $\\Rightarrow$ $\\frac{114}{352}$<\/p>\n Hence option D is the correct answer.<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that $tan\\theta = \\frac{11}{60}$ \u2026 (1)<\/p>\n $\\Rightarrow$ $\\frac{cos\\theta – sin\\theta}{cos\\theta + sin\\theta}$<\/p>\n Dividing both numerator and denominator by $cos\\theta$<\/p>\n $\\Rightarrow$ $\\frac{1 – tan\\theta}{1 + tan\\theta}$<\/p>\n $\\Rightarrow$ $\\frac{1 – \\frac{11}{60}}{1 + \\frac{11}{60}}$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{60-11}{60}}{\\frac{60+11}{60}}$<\/p>\n $\\Rightarrow$ $\\frac{49}{71}$<\/p>\n Hence option A is the correct answer.<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given that $tan \u03b8 + cot \u03b8 = x$<\/p>\n Squaring on both sides<\/p>\n $\\Rightarrow$ $(tan \u03b8 + cot \u03b8)^2 = x^2$<\/p>\n $\\Rightarrow$ $tan^2\u03b8 + cot^2\u03b8+2tan\u03b8\\times cot\u03b8 = x^2$<\/p>\n $\\Rightarrow$ $tan^2\u03b8 + cot^2\u03b8 = x^2 – 2$ (We know that $tan\u03b8\\times cot\u03b8 = 1$)<\/p>\n Squaring on both sides<\/p>\n $\\Rightarrow$ $(tan^2\u03b8 + cot^2\u03b8)^2 = (x^2 – 2)^2$<\/p>\n $\\Rightarrow$ $tan^4\u03b8 + cot^4\u03b8 + 2tan^2\u03b8\\times cot^2\u03b8 = x^4-4x^2+4$<\/p>\n $\\Rightarrow$ $tan^4\u03b8 + cot^4\u03b8 +2=x^4-4x^2+4$ (We know that $tan\u03b8\\times cot\u03b8 = 1$)<\/p>\n $\\Rightarrow$ $tan^4\u03b8 + cot^4\u03b8 =x^2(x^2-4)+4-2$<\/p>\n $\\Rightarrow$ $tan^4\u03b8 + cot^4\u03b8 =x^2(x^2-4)+2$.<\/p>\n Therefore, option B is the right answer.<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that $sin^2\\theta – cos^2\\theta = \\frac{16}{25}$ \u2026 (1)<\/p>\n We know that $sin^2\\theta + cos^2\\theta = 1$ \u2026 (2)<\/p>\n On multiplying both the equations, we get,<\/p>\n $( sin^2\\theta – cos^2\\theta)*(sin^2\\theta + cos^2\\theta) = \\frac{16}{25} * 1$<\/p>\n $\\Rightarrow$ $sin^4\\theta – cos^4\\theta = \\frac{16}{25}$<\/p>\n $\\Rightarrow$ $cos^4\\theta – sin^4\\theta = -\\frac{16}{25}$<\/p>\n Therefore, option C is the right answer.<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that $cosec^2\\theta + cot^2\\theta = \\frac{11}{5}$ \u2026 (1)<\/p>\n We know that $cosec^2\\theta – cot^2\\theta = 1$ \u2026 (2)<\/p>\n On multiplying both the equations, we get,<\/p>\n $\\Rightarrow$ $(cosec^2\\theta + cot^2\\theta)*( cosec^2\\theta – cot^2\\theta) = \\frac{11}{5} * 1$<\/p>\n $\\Rightarrow$ $cosec^4\\theta – cot^4\\theta = \\frac{11}{5}$<\/p>\n Therefore, option A is the right answer.<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n It is given that $tan^2\\theta + sec^2\\theta = \\frac{9}{8}$ \u2026 (1)<\/p>\n We know that $sec^2\\theta – tan^2\\theta = 1$ \u2026 (2)<\/p>\n On multiplying both the equations , we get,<\/p>\n $( tan^2\\theta + sec^2\\theta)*( sec^2\\theta – tan^2\\theta) = \\frac{9}{8} * 1$<\/p>\n $\\Rightarrow$ $sec^4\\theta – tan^4\\theta = \\frac{9}{8}$<\/p>\n $\\Rightarrow$ $tan^4\\theta – sec^4\\theta = \\frac{-9}{8}$<\/p>\n Therefore, option B is the right answer.<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that $cos \u03b8 + sec \u03b8 = a$<\/p>\n Squaring on both sides<\/p>\n $\\Rightarrow$ $(cos \u03b8 + sec \u03b8)^2 = a^2$<\/p>\n $\\Rightarrow$ $cos^2\u03b8 + sec^2\u03b8+2cos\u03b8\\times sec\u03b8 = a^2$<\/p>\n $\\Rightarrow$ $cos^2\u03b8 + sec^2\u03b8 = a^2 – 2$ … (1) (We know that $cos\u03b8\\times sec\u03b8 = 1$)<\/p>\n Taking cube on both sides<\/p>\n $\\Rightarrow$ $(cos^2\u03b8 + sec^2\u03b8)^3 = (a^2 – 2)^3$<\/p>\n $\\Rightarrow$ $cos^6\u03b8 + 3cos^4\u03b8\\times sec^2\u03b8 + 3cos^2\u03b8\\times sec^4\u03b8 + sec^6\u03b8$ = $a^6 – 6a^4 + 12 a^2 – 8$<\/p>\n $\\Rightarrow$ $cos^6\u03b8 + sec^6\u03b8$ = $a^6 – 6a^4 + 12 a^2 – 8$ $- 3cos^2\u03b8\\times sec^2\u03b8 ( cos^2\u03b8 + sec^2\u03b8 )$<\/p>\n $\\Rightarrow$ $cos^6\u03b8 + sec^6\u03b8$ = $a^6 – 6a^4 + 12 a^2 – 8 – 3(a^2 – 2)$ (From equation 1)<\/p>\n $\\Rightarrow$ $cos^6\u03b8 + sec^6\u03b8 = a^6 – 6a^4 + 6a^2 – 2$<\/p>\n Therefore option D is the correct answer.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n In the given diagram PR is the tower and RS is shadow.<\/p>\n $\\Rightarrow$ tan60\u00b0 = $\\frac{PR}{QR}$<\/p>\n $\\Rightarrow$ $\\sqrt{3} = \\frac{PR}{60}$<\/p>\n $\\Rightarrow$ $PR = 60\\sqrt{3}$ m<\/p>\n Therefore the height of the tower is $60\\sqrt{3}$ m. Hence we can say that option B is the correct answer.<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that Squaring on both sides<\/p>\n $\\Rightarrow$ $(tanA-cotA)^2 = 0$<\/p>\n $\\Rightarrow$ $tan^2A+cot^2A-2tanA*cotA = 0$ (We know that $\\tan\\theta\\times \\cot\\theta = 1$)<\/p>\n $\\Rightarrow$ $tan^2A+cot^2A = 2$<\/p>\n Now squaring again on both the sides<\/p>\n $\\Rightarrow$ $(tan^2A+cot^2A)^2 = 2^2$<\/p>\n $\\Rightarrow$ $tan^4A+cot^4A+2tan^2A\\times cot^2A = 4$<\/p>\n $\\Rightarrow$ $tan^4A+cot^4A+2(tanA\\times cotA)^2 = 4$ (We know that $\\tan\\theta\\times \\cot\\theta = 1$)<\/p>\n $\\Rightarrow$ $tan^4A+cot^4A = 4 – 2$<\/p>\n $\\Rightarrow$ $tan^4A+cot^4A = 2$<\/p>\n Therefore option A is the correct answer.<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that $tanA = \\frac{12}{5}$<\/p>\n We have to find out the value of $\\frac{12sinA + 5cosA}{12sinA – 5cosA}$<\/p>\n $\\Rightarrow$ $\\frac{12sinA + 5cosA}{12sinA – 5cosA}$<\/p>\n Dividing numerator and denominator by cosA<\/p>\n $\\Rightarrow$ $\\frac{12(\\frac{sinA}{cosA}) + 5(\\frac{cosA}{cosA})}{12(\\frac{sinA}{cosA}) – 5(\\frac{cosA}{cosA})}$<\/p>\n $\\Rightarrow$ $\\frac{12tanA + 5}{12tanA – 5}$<\/p>\n $\\Rightarrow$ $\\frac{12(\\frac{12}{5}) + 5}{12(\\frac{12}{5}) – 5}$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{144 + 25}{5}}{\\frac{144 – 25}{5}}$<\/p>\n $\\Rightarrow$ $\\frac{169}{119}$<\/p>\n Hence option D is the correct answer.<\/p>\n 21)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that<\/p>\n sinA = $\\frac{-\\sqrt{3}}{2}$<\/p>\n Therefore A = (180\u00b0+ 60\u00b0) or (360\u00b0 – 60\u00b0)<\/p>\n $\\Rightarrow$ A = 240\u00b0 or 300\u00b0<\/p>\n Similarly cosB = $\\frac{\\sqrt{3}}{2}$<\/p>\n Therefore B = 330\u00b0<\/p>\n Hence A+B = 240\u00b0+330\u00b0 or 300\u00b0+330\u00b0<\/p>\n $\\Rightarrow$ (A+B) = 570\u00b0 or 630\u00b0<\/p>\n Hence option D is correct answer.<\/p>\n 22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that<\/p>\n cosA = -$\\frac{1}{\\sqrt{2}}$<\/p>\n Therefore A = 180\u00b0+ 45\u00b0 = 225\u00b0<\/p>\n Similarly cosB = -$\\frac{1}{2}$<\/p>\n Therefore B = 180\u00b0 + 60\u00b0 = 240\u00b0<\/p>\n Hence A+B = 225\u00b0+240 = 465\u00b0<\/p>\n Hence option A is correct answer.<\/p>\n 23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that<\/p>\n sinA = $\\frac{\\sqrt{3}}{2}$<\/p>\n Therefore A = 60\u00b0 or 120\u00b0<\/p>\n Similarly cosB = $\\frac{1}{2}$<\/p>\n Therefore B = 60\u00b0<\/p>\n Hence A+B = 60\u00b0+60\u00b0 or 120\u00b0+60\u00b0<\/p>\n $\\Rightarrow$ (A+B) = 120\u00b0 or 180\u00b0<\/p>\n Hence option A is correct answer.<\/p>\n 24)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given that $\\cos\\theta_{1} = \\frac{\\sqrt{3}}{2}$<\/p>\n We know that $\\sin\\theta = \\sqrt{1-\\cos^2\\theta}$<\/p>\n Therefore $\\sin\\theta_{1} = \\sqrt{1-\\cos^2\\theta_{1}}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{1} = \\sqrt{1-(\\frac{\\sqrt{3}}{2})^2}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{1} = \\frac{1}{2}$<\/p>\n Therefore $\\tan\\theta_{1} = \\frac{\\sin\\theta_{1}}{\\cos\\theta_{1}}$<\/p>\n $\\Rightarrow$ $\\tan\\theta_{1} = \\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $\\Rightarrow$ $\\tan\\theta_{1} = \\frac{1}{\\sqrt{3}}$<\/p>\n Also $\\sin\\theta_{2} = \\sqrt{1-\\cos^2\\theta_{2}}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{2} = \\sqrt{1-(\\frac{24}{25})^2}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{2} = \\frac{7}{25}$<\/p>\n $\\tan\\theta_{2} = \\frac{\\sin\\theta_{2}}{\\cos\\theta_{2}}$<\/p>\n $\\Rightarrow$ $\\tan\\theta_{2} = \\frac{\\frac{7}{25}}{\\frac{24}{25}}$<\/p>\n $\\Rightarrow$ $\\tan\\theta_{2} = \\frac{7}{24}$<\/p>\n We have to find out the value of $\\tan(\\theta_{1}-\\theta_{2})$<\/p>\n $\\Rightarrow$ $\\frac{\\tan\\theta_{1}-\\tan\\theta_{2}}{1+\\tan\\theta_{1}\\times \\tan\\theta_{2}}$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{1}{\\sqrt{3}}-\\frac{7}{24}}{1+\\frac{1}{\\sqrt{3}}\\times \\frac{7}{24}}$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{24 – 7\\sqrt{3}}{24\\sqrt{3}}}{\\frac{24\\sqrt{3} + 7}{24\\sqrt{3}}}$<\/p>\n $\\Rightarrow$ $\\frac{24 – 7\\sqrt{3}}{24\\sqrt{3} + 7}$<\/p>\n Hence option C is the correct answer.<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that $\\cos\\theta_{1} = \\frac{1}{\\sqrt{2}}$<\/p>\n We know that $\\sin\\theta = \\sqrt{1-\\cos^2\\theta}$<\/p>\n Therefore $\\sin\\theta_{1} = \\sqrt{1-\\cos^2\\theta_{1}}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{1} = \\sqrt{1-(\\frac{1}{\\sqrt{2}})^2}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{1} = \\frac{1}{\\sqrt{2}}$<\/p>\n Also $\\sin\\theta_{2} = \\sqrt{1-\\cos^2\\theta_{2}}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{2} = \\sqrt{1-(\\frac{40}{41})^2}$<\/p>\n $\\Rightarrow$ $\\sin\\theta_{2} = \\frac{9}{41}$<\/p>\n We have to find out the value of $\\cos(\\theta_{1}-\\theta_{2})$<\/p>\n $\\Rightarrow$ $\\cos\\theta_{1}\\cos\\theta_{2} + \\sin\\theta_{1}\\sin\\theta_{2}$<\/p>\n $\\Rightarrow$ $\\frac{1}{\\sqrt{2}}\\times \\frac{40}{41} + \\frac{1}{\\sqrt{2}}\\times \\frac{9}{41}$<\/p>\n $\\Rightarrow$ $\\frac{40 + 9}{41\\sqrt{2}}$<\/p>\n $\\Rightarrow$ $\\frac{49\\sqrt{2}}{82}$<\/p>\n Hence option A is the correct answer.<\/p>\n 26)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given that $\\sin\\theta_{1} = \\frac{3}{5}$ We have to find out the value of $\\sin(\\theta_{1}+\\theta_{2})$ Hence option B is the correct answer.<\/p>\n 27)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{\\sin x}{1 + \\cos x} + \\frac{1 +\\ cos x}{\\sin x} = 4$ =>$2\\cosec x = 4$ 28)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\cos 4x = \\sin x$ 29)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $2a + a^2\\tan 2x = \\tan 2x$ 30)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We are given that $\\cos\\theta = \\frac{3}{5}$<\/p>\n $\\Rightarrow$ $\\sin^2\\theta = \\sqrt{1 – \\cos^2\\theta}$<\/p>\n $\\Rightarrow$ $\\sin^2\\theta = \\sqrt{1 – (\\frac{3}{5})^2}$<\/p>\n $\\Rightarrow$ $\\sin^2\\theta = \\frac{16}{25}$<\/p>\n $\\Rightarrow$ $\\sin\\theta = \\frac{4}{5}$ ($0 \\leq \\theta \\leq 90$)<\/p>\n Therefore $\\tan\\theta = \\frac{\\sin\\theta}{cos\\theta}$<\/p>\n $\\Rightarrow$ $\\tan\\theta = \\frac{\\frac{4}{5}}{\\frac{3}{5}}$<\/p>\n $\\Rightarrow$ $\\tan\\theta = \\frac{4}{3}$ … (1)<\/p>\n We can easily start $\\cot\\theta=\\frac{3}{4}$ … (2)<\/p>\n We have to find out, $\\frac{\\frac{4}{3} – \\frac{3}{4}}{\\frac{4}{3} + \\frac{3}{4} }$<\/p>\n $\\Rightarrow$ $\\frac{16 – 9}{16 + 9}$<\/p>\n $\\Rightarrow$ $\\frac{7}{25}$<\/p>\n Hence option C is the correct answer.<\/p>\n 31)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that $\\frac{\\tan^2\\theta+3}{\\tan^2\\theta-3} = 2$.<\/p>\n Assuming $\\tan^2\\theta = t $, we get,<\/p>\n $\\frac{t + 3}{t – 3} = 2$<\/p>\n $\\Rightarrow$ $t = 9$<\/p>\n $\\Rightarrow$ $\\tan^2\\theta=9$<\/p>\n $\\Rightarrow$ $\\tan\\theta = 3$ ($\\theta$ is in 1st quadrant)<\/p>\n We know that $\\sec^2\\theta – \\tan^2\\theta = 1$<\/p>\n $\\Rightarrow$ $\\sec^2\\theta = 1 + \\tan^2\\theta $<\/p>\n Also $\\cot^2\\theta = \\frac{1}{\\tan^2\\theta}$<\/p>\n We have to find out $\\frac{\\sec^2\\theta+\\cot^2\\theta}{\\sec^2\\theta-\\cot^2\\theta}$<\/p>\n $\\Rightarrow$ $\\frac{1 + \\tan^2\\theta+\\frac{1}{\\tan^2\\theta}}{1 + \\tan^2\\theta – \\frac{1}{\\tan^2\\theta}}$<\/p>\n $\\Rightarrow$ $\\frac{1 + 9 +\\frac{1}{9}}{1 + 9 – \\frac{1}{9}}$<\/p>\n $\\Rightarrow$ $\\frac{91}{89}$<\/p>\n Hence option D is the correct answer.<\/p>\n 32)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that $cos\\theta = \\frac{\\sqrt{2+\\sqrt{3}}}{2}$<\/p>\n Then,<\/p>\n $\\Rightarrow$ $cos\\theta +sec\\theta$<\/p>\n $\\Rightarrow$ $cos\\theta +\\frac{1}{cos\\theta}$<\/p>\n $\\Rightarrow$ $\\frac{cos^2\\theta +1}{cos\\theta}$<\/p>\n $\\Rightarrow$ $\\frac{(\\frac{\\sqrt{2+\\sqrt{3}}}{2})^2 + 1}{\\frac{\\sqrt{2+\\sqrt{3}}}{2}}$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{2+\\sqrt{3}}{4}+ 1}{\\frac{\\sqrt{2+\\sqrt{3}}}{2}}$<\/p>\n $\\Rightarrow$ $\\frac{6 +\\sqrt{3}}{2\\sqrt{2+\\sqrt{3}}}$<\/p>\n 33)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given that $\\frac{\\sec^2\\theta+1}{\\sec^2\\theta-1} = \\frac{37}{35}$.<\/p>\n Assuming $\\sec^2\\theta = t $, we get,<\/p>\n $\\frac{t + 1}{t – 1} = \\frac{37}{35}$<\/p>\n $\\Rightarrow$ $t = 36$<\/p>\n $\\Rightarrow$ $\\sec^2\\theta=36$<\/p>\n $\\Rightarrow$ $\\sec\\theta = 6$ ($\\theta$ is in 1st quadrant)<\/p>\n We have to find out value of $\\frac{\\tan\\theta+\\sin\\theta}{\\tan\\theta-\\sin\\theta}$.<\/p>\n $\\frac{\\tan\\theta+\\sin\\theta}{\\tan\\theta-\\sin\\theta}$ = $\\frac{\\sec\\theta+1}{\\sec\\theta-1}$ (Dividing numerator and denominator by $\\sin\\theta$)<\/p>\n $\\Rightarrow$ $\\frac{6 + 1}{6 – 1}$<\/p>\n $\\Rightarrow$ $\\frac{7}{5}$.<\/p>\n Therefore, option B is the right answer.<\/p>\n 34)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let us assume that height of flag is h meters.<\/p>\n In triangle AEF, In triangle ABC, 35)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that $\\cos\\theta – \\sec\\theta = 7$ 36)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given that $\\sin\\theta + \\cosec\\theta = 2$ (We know that $\\cosec\\theta = \\frac{1}{\\sin\\theta}$) 37)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given that $\\sec\\theta$ = $\\frac{15}{9}$<\/p>\n $\\cos\\theta$ = $\\frac{9}{15}$<\/p>\n $\\Rightarrow$ $\\sin\\theta = \\sqrt{1- \\cos^2\\theta}$<\/p>\n $\\Rightarrow$ $\\sin\\theta = \\sqrt{1- (\\frac{9}{15})^2}$<\/p>\n $\\Rightarrow$ $\\sin\\theta = \\frac{12}{15}$<\/p>\n $\\Rightarrow$ $\\cosec\\theta = \\frac{15}{12}$ (Inverse of $\\sin\\theta$)<\/p>\n $\\Rightarrow$ $\\cot\\theta = \\frac{\\cos\\theta}{\\sin\\theta}$<\/p>\n $\\Rightarrow$ $\\cot\\theta = \\frac{\\frac{9}{15}}{\\frac{12}{15}}$<\/p>\n $\\Rightarrow$ $\\cot\\theta = \\frac{9}{12}$<\/p>\n We have to find<\/p>\n $\\Rightarrow$ $\\frac{\\cot\\theta + \\cosec\\theta}{\\cot\\theta – \\cosec\\theta }$<\/p>\n $\\Rightarrow$ $\\frac{\\frac{9}{12} + \\frac{15}{12}}{\\frac{9}{12} – \\frac{15}{12}}$<\/p>\n $\\Rightarrow$ $\\frac{9 + 15 }{9 – 15}$<\/p>\n $\\Rightarrow$ $-4$ 38)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given that: $sec\\theta=\\frac{25}{24}$<\/p>\n $cos\\theta=\\frac{24}{25}$<\/p>\n Using, $sin^2\\theta+cos^2\\theta =1$<\/p>\n $\\Rightarrow sin^2\\theta=1-(\\frac{24}{25})^2= \\frac{625-576}{625}$<\/p>\n $\\Rightarrow sin^2\\theta=\\frac{49}{625}$<\/p>\n $\\Rightarrow sin\\theta=\\frac{7}{25}$<\/p>\n We know that $cot\\theta = \\frac{cos\\theta}{sin\\theta}$<\/p>\n $\\Rightarrow cot \\theta =\\frac{\\frac{24}{25}}{\\frac{7}{25}}$<\/p>\n $\\Rightarrow cot \\theta =\\frac{24}{7} $.<\/p>\n Therefore, option C is the right answer.<\/p>\n 39)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given that cos \u03b8 + sec \u03b8 = a Therefore, option A is the right answer.<\/p>\n 40)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $x + y = z$<\/p>\n $tan z = tan (x + y)$ = $\\frac{tan x + tan y}{1 – tan x * tan y}$<\/p>\n $\\Rightarrow$ $tan z = \\frac{m}{1 -n}$<\/p>\n $\\Rightarrow$ $\\frac{m}{1 -n}$ = 20<\/p>\n Hence, option B is correct.<\/p>\n
\n(Assume that 180\u00b0 $\\leq$ A, B $\\leq$ 360\u00b0)<\/p>\n
\n(Assume that 180\u00b0 $\\leq$ A, B $\\leq$ 270\u00b0)<\/p>\n
\n(Assume that 0\u00b0 $\\leq$ A, B $\\leq$ 180\u00b0)<\/p>\n
\nWhat is the value of $\\cot x$<\/p>\n
\nWhat is the value of $\\sec^2x$?<\/p>\n
\n$\\therefore AC=\\sqrt{2}\\star BC $
\nsinA=$\\frac{BC}{AC}$<\/p>\n
\n$\\sin^{2}\\theta+1=(\\frac{5}{2})\\sin\\theta$
\n$2\\sin^{2}\\theta-5\\sin\\theta+1=0$
\n$(\\sin\\theta-2)(2\\sin\\theta-1)=0$
\nTherefore $\\sin\\theta=\\frac{1}{2}$
\n$\\therefore \\theta=30$
\n$\\tan30=\\frac{1}{2}$.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/pp_fPOcCKc.PNG\")
<\/figure>\n
\n$\\Rightarrow$ $tanA-cotA =0$<\/p>\n
\n$5tanA = 12$<\/p>\n
\nWe know that $\\cos\\theta = \\sqrt{1-\\sin^2\\theta}$
\nTherefore $\\cos\\theta_{1} = \\sqrt{1-\\sin^2\\theta_{1}}$
\n$\\Rightarrow$ $\\cos\\theta_{1} = \\sqrt{1-(\\frac{3}{5})^2}$
\n$\\Rightarrow$ $\\cos\\theta_{1} = \\frac{4}{5}$
\nAlso $\\sin\\theta = \\sqrt{1-\\cos^2\\theta}$
\nTherefore $\\sin\\theta_{2} = \\sqrt{1-\\cos^2\\theta_{2}}$
\n$\\Rightarrow$ $\\sin\\theta_{2} = \\sqrt{1-(\\frac{7}{25})^2}$
\n$\\Rightarrow$ $\\sin\\theta_{2} = \\frac{24}{25}$<\/p>\n
\n$\\Rightarrow$ $\\sin\\theta_{1}\\cos\\theta_{2} + \\cos\\theta_{1}\\sin\\theta_{2}$
\n$\\Rightarrow$ $\\frac{3}{5}\\times \\frac{7}{25} + \\frac{4}{5}\\times \\frac{24}{25}$
\n$\\Rightarrow$ $\\frac{21 + 96}{125}$
\n$\\Rightarrow$ $\\frac{117}{125}$<\/p>\n
\n=>$\\frac{\\sin^2x + (1 + \\cos x)^2}{\\sin x(1 + \\cos x)} = 4$
\n=>$\\frac{\\sin^2x + 1 + \\cos^2x + 2\\cos x}{\\sin x(1 + \\cos x)} = 4$
\n=>$\\frac{2 + 2\\cos x}{\\sin x(1 + \\cos x)} = 4$
\n=>$\\frac{2 + 2\\cos x}{\\sin x(1 + \\cos x)} = 4$
\n=>$\\frac{2(1 + \\cos x)}{\\sin x(1 + \\cos x)} = 4$<\/p>\n
\n=>$\\cosec x = 2$
\n=>$\\sin x = \\frac{1}{2}$
\n=> $x = 30$\u00b0
\nso, $\\cot x = \\cot 30$\u00b0 = $\\sqrt{3}$
\nHence, option B is the correct answer.<\/p>\n
\n=>$\\cos 4x = \\cos(90$\u00b0$ – x)$
\n=>$4x = 90$\u00b0$ – x$
\n=>$x = 18$\u00b0$$
\nSo, $\\sin 5x = \\sin 90$\u00b0$ = 1$
\nHence, option A is the correct answer.<\/p>\n
\n=>$2a = \\tan 2x – a^2\\tan 2x$
\n=>$\\tan 2x = \\frac{2a}{1 – a^2}$
\nwe know that $\\tan 2x$ = $\\frac{2\\tan x}{1 – \\tan^2x}$
\nOn comparing we get $\\tan x = a$
\n$\\sec^2x = 1 + \\tan^2x = 1 + a^2$
\nHence, option B is the correct answer.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/ABC.PNG\")
<\/figure>\n
\ntan(45) = $\\frac{AE}{EF}$ = $1$
\nAE =EF =600 m
\nWe can see that BC = EF
\nHence BC = 600m … (1)<\/p>\n
\ntan(30) = $\\frac{AB}{BC}$ = $\\frac{1}{\\sqrt{3}}$
\ntan(30) = $\\frac{600-h}{BC}$ =$\\frac{1}{\\sqrt{3}}$
\n$\\Rightarrow$ $\\frac{600-h}{600}$ =$\\frac{1}{\\sqrt{3}}$
\n$\\Rightarrow$ $h$ =$600 – 200\\sqrt{3}$
\n$\\Rightarrow$ $h$ =$200\\sqrt{3}(\\sqrt{3} – 1)$
\nHence, option C is the right choice.<\/p>\n
\nSquaring on both sides
\n$\\Rightarrow$ $(\\cos\\theta – \\sec\\theta)^2 = 7^2$
\n$\\Rightarrow$ $\\cos^2\\theta – 2\\times\\cos\\theta\\sec\\theta + \\sec^2\\theta = 49$ (We know that $\\cos\\theta\\sec\\theta = 1$)
\n$\\Rightarrow$ $\\cos^2\\theta – 2\\times1 + \\sec^2\\theta = 49$
\n$\\Rightarrow$ $\\cos^2\\theta + \\sec^2\\theta = 49 + 2$
\n$\\Rightarrow$ $\\cos^2\\theta + \\sec^2\\theta = 51$
\nTherefore, option A is the right answer.<\/p>\n
\n$\\Rightarrow$ $\\sin\\theta + \\frac{1}{\\sin\\theta}= 2$
\n$\\Rightarrow$ $\\sin^2\\theta – 2\\times\\sin\\theta + 1 = 0$
\n$\\Rightarrow$ $(\\sin\\theta – 1)^2 = 0$
\n$\\therefore$ $\\sin\\theta = +1 or -1$
\nSince $\\sin\\theta = +1 or -1$, $\\cosec\\theta$ is also equal to $-1 or +1$.
\nHence, the value of $\\cosec^{32}\\theta + \\sin^{32}\\theta$ = $1 + 1$ = $2$.
\nTherefore, option D is the right answer.<\/p>\n
\nTherefore, option C is the right answer.<\/p>\n
\nSquaring both sides,
\n$\\Rightarrow$ $(cos \u03b8 + sec \u03b8)^2=(a)^2$
\n$\\Rightarrow$ $cos^2\\theta+sec^2\\theta+2(cos\\theta)(sec\\theta)=a^2$
\n$\\Rightarrow$ $cos^2\\theta+sec^2\\theta+2=a^2$ [$\\because cos\\theta sec\\theta=1$]
\n$\\Rightarrow$ $cos^2\\theta+sec^2\\theta=a^2-2$
\nAgain squaring both sides, we get :
\n$\\Rightarrow$ $(cos^2\\theta+sec^2\\theta)^2=(a^2-2)^2$
\n$\\Rightarrow$ $cos^4\\theta+sec^4\\theta+2(cos^2\\theta)(sec^2\\theta)=a^4-4a^2+4$
\n$\\Rightarrow$ $cos^4\\theta+sec^4\\theta+2=a^4-4a^2+4$
\n$\\Rightarrow$ $cos^4\\theta+sec^4\\theta=a^2(a^2-4)+4-2$
\n$\\Rightarrow$ $cos^4\\theta+sec^4\\theta=a^2(a^2-4)+2$.<\/p>\n