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Download Trigonometry Questions for SSC MTS<\/a><\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>If $\\cot \\theta = \\frac{15}{8}, \\theta$ is an acute angle, then find the value of $\\frac{(1 – \\cos \\theta)(2 + 2 \\cos \\theta)}{(2 – 2 \\sin \\theta)(1 + \\sin \\theta)}$.<\/p>\n a)\u00a0$\\frac{225}{64}$<\/p>\n b)\u00a0$\\frac{64}{225}$<\/p>\n c)\u00a0$\\frac{16}{15}$<\/p>\n d)\u00a0$\\frac{8}{15}$<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\cot\\theta=\\frac{15}{8}$<\/p>\n $\\tan\\theta=\\frac{8}{15}$<\/p>\n $\\sec\\theta=\\sqrt{1+\\tan^2\\theta\\ }=\\sqrt{1+\\left(\\frac{8}{15}\\right)^2}=\\sqrt{\\frac{289}{225}\\ }=\\frac{17}{15}$<\/p>\n $\\cos\\theta=\\frac{15}{17}$<\/p>\n $\\sin\\theta=\\sqrt{1-\\cos^2\\theta\\ }=\\sqrt{1-\\left(\\frac{15}{17}\\right)^2}=\\sqrt{1-\\frac{225}{289}}=\\sqrt{\\frac{64}{289}}=\\frac{8}{17}$<\/p>\n $\\frac{(1-\\cos\\theta)(2+2\\cos\\theta)}{(2-2\\sin\\theta)(1+\\sin\\theta)}=\\frac{\\left(1-\\frac{15}{17}\\right)\\left[2+2\\left(\\frac{15}{17}\\right)\\right]}{\\left[2-2\\left(\\frac{8}{17}\\right)\\right]\\left(1+\\frac{8}{17}\\right)}$<\/p>\n $=\\frac{\\left(\\frac{2}{17}\\right)\\left[\\frac{64}{17}\\right]}{\\left[\\frac{18}{17}\\right]\\left(\\frac{25}{17}\\right)}$<\/p>\n $=\\frac{2\\times64}{18\\times25}$<\/p>\n $=\\frac{64}{225}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 2:\u00a0<\/b>If $2 \\cos^2 \\theta – 5 \\cos \\theta + 2 = 0, 0^\\circ < \\theta < 90^\\circ$, then the value of $(\\sec \\theta + \\tan \\theta)$ is:<\/p>\n a)\u00a0$2 + \\sqrt{3}$<\/p>\n b)\u00a0$1 – \\sqrt{3}$<\/p>\n c)\u00a0$1 + \\sqrt{3}$<\/p>\n d)\u00a0$2 – \\sqrt{3}$<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2\\cos^2\\theta-5\\cos\\theta+2=0$<\/p>\n $2\\cos^2\\theta-4\\cos\\theta-\\cos\\theta+2=0$<\/p>\n $2\\cos\\theta\\left(\\cos\\theta-2\\right)-1\\left(\\cos\\theta-2\\right)=0$<\/p>\n $\\left(\\cos\\theta-2\\right)\\left(2\\cos\\theta-1\\right)=0$<\/p>\n $\\cos\\theta=2$\u00a0 or\u00a0$\\cos\\theta=\\frac{1}{2}$<\/p>\n $\\cos\\theta=2$ is not possible.<\/p>\n So, $\\cos\\theta=\\frac{1}{2}$<\/p>\n $\\sec\\theta=2$<\/p>\n $\\tan\\theta=\\sqrt{\\sec^2\\theta-1}=\\sqrt{4-1}=\\sqrt{3}$<\/p>\n $\\therefore$\u00a0\u00a0$\\left(\\sec\\theta+\\tan\\theta\\right)=2+\\sqrt{3}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>Find the value of $\\tan35^{\\circ}\\cot40^{\\circ}\\tan45^{\\circ}\\cot50^{\\circ}\\tan55^{\\circ}$.<\/p>\n a)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a0-1<\/p>\n d)\u00a01<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\tan35^{\\circ}\\cot40^{\\circ}\\tan45^{\\circ}\\cot50^{\\circ}\\tan55^{\\circ}=\\tan35^{\\circ}\\cot40^{\\circ}\\tan45^{\\circ}\\cot\\left(90-40^{\\circ}\\right)\\tan\\left(90-35^{\\circ}\\right)$<\/p>\n $=\\tan35^{\\circ}\\cot40^{\\circ}\\left(1\\right)\\tan40^{\\circ}\\cot35^{\\circ}$<\/p>\n $=1$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 4:\u00a0<\/b>If $\\sin(20 + x)^\\circ = \\cos 60^\\circ, 0 \\leq (20 + x) \\leq 90$, then find the value of $2 \\sin^2(3x + 15)^\\circ – \\cosec^2(2x + 10)^\\circ$.<\/p>\n a)\u00a03<\/p>\n b)\u00a0-3<\/p>\n c)\u00a0$-\\frac{1}{3}$<\/p>\n d)\u00a0-2<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin(20+x)^{\\circ}=\\cos60^{\\circ}$<\/p>\n $\\sin(20+x)^{\\circ}=\\sin30^{\\circ}$<\/p>\n $20+x=30$<\/p>\n $x=10$<\/p>\n $2\\sin^2(3x+15)^{\\circ}-\\operatorname{cosec}^2(2x+10)^{\\circ}=2\\sin^245^{\\circ}-\\operatorname{cosec}^230^{\\circ}$<\/p>\n $=2\\left(\\frac{1}{\\sqrt{2}}\\right)^2-\\left(2\\right)^2$<\/p>\n $=2\\left(\\frac{1}{2}\\right)-4$<\/p>\n $=1-4$<\/p>\n $=-3$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>If $\\sin^6 \\theta + \\cos^6 \\theta = \\frac{1}{3}, 0^\\circ < \\theta < 90^\\circ$, then what is the value of $\\sin \\theta \\cos \\theta$?<\/p>\n a)\u00a0$\\frac{\\sqrt{2}}{3}$<\/p>\n b)\u00a0$\\frac{\\sqrt{6}}{6}$<\/p>\n c)\u00a0$\\frac{\\sqrt{2}}{\\sqrt{3}}$<\/p>\n d)\u00a0$\\frac{2}{9}$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin^6\\theta+\\cos^6\\theta=\\frac{1}{3}$<\/p>\n $\\left(\\sin^2\\theta\\right)^3+\\left(\\cos^2\\theta\\right)^3=\\frac{1}{3}$<\/p>\n $\\left(\\sin^2\\theta+\\cos^2\\theta\\right)\\left(\\sin^4\\theta-\\sin^2\\theta\\ \\cos^2\\theta+\\cos^4\\theta\\ \\right)=\\frac{1}{3}$<\/p>\n $\\left(1\\right)\\left(\\sin^4\\theta+\\cos^4\\theta+2\\sin^2\\theta\\ \\cos^2\\theta-3\\sin^2\\theta\\ \\cos^2\\theta\\ \\right)=\\frac{1}{3}$<\/p>\n $\\left(\\sin^2\\theta+\\cos^2\\theta\\right)^2-3\\sin^2\\theta\\ \\cos^2\\theta\\ =\\frac{1}{3}$<\/p>\n $1-3\\sin^2\\theta\\ \\cos^2\\theta\\ =\\frac{1}{3}$<\/p>\n $3\\sin^2\\theta\\ \\cos^2\\theta\\ =\\frac{2}{3}$<\/p>\n $\\sin^2\\theta\\ \\cos^2\\theta\\ =\\frac{2}{9}$<\/p>\n $\\sin\\theta\\ \\cos\\theta=\\frac{\\sqrt{2}}{3}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>If $\\sin(\\theta+30^{\\circ})=\\frac{3}{\\sqrt{12}}$, then the value of $\\theta$ is equal to:<\/p>\n a)\u00a0$15^\\circ$<\/p>\n b)\u00a0$60^\\circ$<\/p>\n c)\u00a0$30^\\circ$<\/p>\n d)\u00a0$45^\\circ$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0$\\sin(\\theta+30^{\\circ})=\\frac{3}{\\sqrt{12}}$<\/p>\n $\\Rightarrow$ \u00a0$\\sin(\\theta+30^{\\circ})=\\frac{3}{2\\sqrt{3}}$<\/p>\n $\\Rightarrow$ \u00a0$\\sin(\\theta+30^{\\circ})=\\frac{\\sqrt{3}\\times\\sqrt{3}}{2\\sqrt{3}}$<\/p>\n $\\Rightarrow$ \u00a0$\\sin(\\theta+30^{\\circ})=\\frac{\\sqrt{3}}{2}$<\/p>\n $\\Rightarrow$ \u00a0$\\sin(\\theta+30^{\\circ})=\\sin60^{\\circ\\ }$<\/p>\n $\\Rightarrow$ \u00a0$\\theta+30^{\\circ}=60^{\\circ\\ }$<\/p>\n $\\Rightarrow$ \u00a0$\\theta=30^{\\circ\\ }$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 7:\u00a0<\/b>If $\\frac{\\sec\\theta+\\tan\\theta}{\\sec\\theta-\\tan\\theta}=5$ and $\\theta$ is an acute angle, then the value of $\\frac{3\\cos^2\\theta+1}{3\\cos^2\\theta-1}$ is:<\/p>\n a)\u00a04<\/p>\n b)\u00a03<\/p>\n c)\u00a01<\/p>\n d)\u00a02<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0$\\frac{\\sec\\theta+\\tan\\theta}{\\sec\\theta-\\tan\\theta}=5$<\/p>\n $\\Rightarrow$ \u00a0$\\sec\\theta+\\tan\\theta=5\\sec\\theta\\ -5\\tan\\theta\\ $<\/p>\n $\\Rightarrow$ \u00a0$6\\tan\\theta=4\\sec\\theta$<\/p>\n $\\Rightarrow$ \u00a0$6\\frac{\\sin\\theta\\ }{\\cos\\theta\\ }=4\\frac{1}{\\cos\\theta\\ }$<\/p>\n $\\Rightarrow$ \u00a0$\\sin\\theta=\\frac{2}{3}$<\/p>\n $\\cos^2\\theta\\ =1-\\sin^2\\theta\\ =1-\\left(\\frac{2}{3}\\right)^2=\\frac{5}{9}$<\/p>\n $\\therefore\\ $ $\\frac{3\\cos^2\\theta+1}{3\\cos^2\\theta-1}=\\frac{3\\left(\\frac{5}{9}\\right)+1}{3\\left(\\frac{5}{9}\\right)-1}$<\/p>\n $=\\frac{\\frac{5}{3}+1}{\\frac{5}{3}-1}$<\/p>\n $=\\frac{8}{2}$<\/p>\n $=4$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 8:\u00a0<\/b>If sin(A – B) = $\\frac{1}{2}$ and cos(A + B) = $\\frac{1}{2}$ where A > B > 0$^\\circ$ and A + B is an acute angle, then the value of A is:<\/p>\n a)\u00a0$30^\\circ$<\/p>\n b)\u00a0$60^\\circ$<\/p>\n c)\u00a0$45^\\circ$<\/p>\n d)\u00a0$15^\\circ$<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0sin(A – B) = $\\frac{1}{2}$<\/p>\n $\\Rightarrow$ \u00a0sin(A – B) = sin 30$^{\\circ\\ }$<\/p>\n $\\Rightarrow$\u00a0 A – B = 30$^{\\circ\\ }$ ……..(1)<\/p>\n cos(A + B) = $\\frac{1}{2}$ and A + B is an acute angle<\/p>\n $\\Rightarrow$ \u00a0cos(A + B) = cos 60$^{\\circ\\ }$<\/p>\n $\\Rightarrow$\u00a0 A + B = 60$^{\\circ\\ }$ ……..(2)<\/p>\n Adding equations (1) and (2),<\/p>\n 2A = 90$^{\\circ\\ }$<\/p>\n $\\Rightarrow$\u00a0 A = 45$^{\\circ\\ }$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 9:\u00a0<\/b>If $2x=\\sin\\theta$ and $\\frac{2}{x}=\\cos\\theta$, then the value of $4\\left(x^2+\\frac{1}{x^2}\\right)$ is:<\/p>\n a)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a04<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$2x=\\sin\\theta$ and\u00a0$\\frac{2}{x}=\\cos\\theta$<\/p>\n From the trigonometric identities,<\/p>\n $\\sin^2\\theta\\ +\\cos^2\\theta\\ =1$<\/p>\n $\\Rightarrow$ $\\left(2x\\right)^2+\\left(\\frac{2}{x}\\right)^2=1$<\/p>\n $\\Rightarrow$\u00a0 $4x^2+\\frac{4}{x^2}=1$<\/p>\n $\\Rightarrow$\u00a0 $4\\left(x^2+\\frac{1}{x^2}\\right)=1$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 10:\u00a0<\/b>If X = $\\tan 40^\\circ$, then the value of $2 \\tan 50^\\circ$ will be:<\/p>\n a)\u00a0$\\frac{2}{X}$<\/p>\n b)\u00a02X<\/p>\n c)\u00a0$\\frac{1}{X}$<\/p>\n d)\u00a0$\\frac{1}{2X}$<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0X = $\\tan 40^\\circ$<\/p>\n $\\therefore\\ 2\\tan50^{\\circ}=2\\tan\\left(90-40^{\\circ}\\right)$<\/p>\n $=2\\cot40^{\\circ}$<\/p>\n $=\\frac{2}{\\tan40^{\\circ\\ }}$<\/p>\n $=\\frac{2}{X\\ }$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 11:\u00a0<\/b>If $2\\cot\\theta=3$, then $\\frac{\\sqrt{13}\\cos\\theta-3\\tan\\theta}{3\\tan\\theta+\\sqrt{13}\\sin\\theta}$ is:<\/p>\n a)\u00a0$\\frac{3}{4}$<\/p>\n b)\u00a0$\\frac{1}{4}$<\/p>\n c)\u00a0$\\frac{2}{3}$<\/p>\n d)\u00a0$\\frac{1}{5}$<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, $2\\cot\\theta=3$<\/p>\n $\\Rightarrow$ \u00a0$\\cot\\theta=\\frac{3}{2}$<\/p>\n $\\tan\\theta=\\frac{2}{3}$<\/p>\n $\\operatorname{cosec}\\theta\\ =\\sqrt{1+\\cot^2\\theta\\ }=\\sqrt{1+\\left(\\frac{3}{2}\\right)2}=\\sqrt{1+\\frac{9}{4}}=\\sqrt{\\frac{13}{4}}=\\frac{\\sqrt{13}}{2}$<\/p>\n $\\sin\\theta\\ =\\frac{2}{\\sqrt{13}}$<\/p>\n $\\cos\\theta\\ =\\sqrt{1-\\sin^2\\theta\\ }=\\sqrt{1-\\frac{4}{13}}=\\sqrt{1-\\left(\\frac{2}{\\sqrt{13}}\\right)^2}=\\sqrt{\\frac{9}{13}}=\\frac{3}{\\sqrt{13}}$<\/p>\n $\\therefore\\ $\u00a0$\\frac{\\sqrt{13}\\cos\\theta-3\\tan\\theta}{3\\tan\\theta+\\sqrt{13}\\sin\\theta}=\\frac{\\sqrt{13}\\left(\\frac{3}{\\sqrt{13}}\\right)-3\\left(\\frac{2}{3}\\right)}{3\\left(\\frac{2}{3}\\right)+\\sqrt{13}\\left(\\frac{2}{\\sqrt{13}}\\right)}$<\/p>\n $=\\frac{3-2}{2+2}$<\/p>\n $=\\frac{1}{4}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 12:\u00a0<\/b>If\u00a0 $x\\sin^3\\theta+y\\cos^3\\theta=\\sin\\theta\\cos\\theta$ and\u00a0 $x\\sin\\theta=y\\cos\\theta$, then the value of $x^2 + y^2$ is:<\/p>\n a)\u00a04<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a01<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$x\\sin\\theta=y\\cos\\theta$<\/p>\n $\\Rightarrow$ \u00a0$y=\\frac{x\\sin\\theta}{\\cos\\theta\\ }$<\/p>\n $x\\sin^3\\theta+y\\cos^3\\theta=\\sin\\theta\\cos\\theta$<\/p>\n $\\Rightarrow$ \u00a0$x\\sin^3\\theta+\\frac{x\\sin\\theta\\ }{\\cos\\theta\\ }\\cos^3\\theta=\\sin\\theta\\cos\\theta$<\/p>\n $\\Rightarrow$ \u00a0$x\\sin^3\\theta+x\\sin\\theta\\ \\cos^2\\theta=\\sin\\theta\\cos\\theta$<\/p>\n $\\Rightarrow$ \u00a0$x\\sin\\theta\\ \\left(\\sin^2\\theta+\\cos^2\\theta\\right)=\\sin\\theta\\cos\\theta$<\/p>\n $\\Rightarrow$ \u00a0$x\\sin\\theta\\ =\\sin\\theta\\cos\\theta$<\/p>\n $\\Rightarrow$ \u00a0$x=\\cos\\theta$<\/p>\n $\\therefore\\ $ $y=\\frac{x\\sin\\theta}{\\cos\\theta\\ }=\\frac{\\cos\\theta\\ \\sin\\theta}{\\cos\\theta\\ }$<\/p>\n $\\Rightarrow$ \u00a0$y=\\sin\\theta\\ $<\/p>\n $\\therefore\\ $ $x^2+y^2=\\cos^2\\theta\\ +\\sin^2\\theta=1\\ $<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 13:\u00a0<\/b>If $\\sqrt{3}\\cos\\theta=\\sin\\theta$, then the value of $\\frac{4\\sin^2\\theta-5\\cos\\theta}{3\\cos\\theta+1}$ is:<\/p>\n a)\u00a0$\\frac{1}{4}$<\/p>\n b)\u00a0$\\frac{1}{5}$<\/p>\n c)\u00a05<\/p>\n d)\u00a0$\\frac{2}{5}$<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0$\\sqrt{3}\\cos\\theta=\\sin\\theta$<\/p>\n $\\Rightarrow$ \u00a0$\\sqrt{3}=\\frac{\\sin\\theta}{\\cos\\theta\\ }$<\/p>\n $\\Rightarrow$\u00a0 $\\tan\\theta\\ =\\sqrt{3}$<\/p>\n $\\Rightarrow$ \u00a0$\\tan\\theta\\ =\\tan60^{\\circ\\ }$<\/p>\n $\\Rightarrow$ \u00a0$\\theta =60^{\\circ\\ }$<\/p>\n $\\therefore\\ $ $\\frac{4\\sin^2\\theta-5\\cos\\theta}{3\\cos\\theta+1}=\\frac{4\\sin^260^{\\circ\\ }-5\\cos60^{\\circ\\ }}{3\\cos60^{\\circ}+1}$<\/p>\n $=\\frac{4\\left(\\frac{\\sqrt{3}}{2}\\right)^2-5\\left(\\frac{1}{2}\\right)}{3\\left(\\frac{1}{2}\\right)+1}$<\/p>\n $=\\frac{4\\left(\\frac{3}{4}\\right)-\\frac{5}{2}}{\\frac{3}{2}+1}$<\/p>\n $=\\frac{3-\\frac{5}{2}}{\\frac{5}{2}}$<\/p>\n $=\\frac{\\frac{1}{2}}{\\frac{5}{2}}$<\/p>\n $=\\frac{1}{5}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>If $\\cos3\\theta=\\sin(\\theta-34^{\\circ})$, then the value of $\\theta$ as an acute angle is:<\/p>\n a)\u00a0$56^\\circ$<\/p>\n b)\u00a0$17^\\circ$<\/p>\n c)\u00a0$31^\\circ$<\/p>\n d)\u00a0$34^\\circ$<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0$\\cos3\\theta=\\sin(\\theta-34^{\\circ})$<\/p>\n $\\Rightarrow$ \u00a0$\\cos3\\theta=\\cos\\left[90-(\\theta-34^{\\circ})\\right]$<\/p>\n $\\Rightarrow$ \u00a0$\\cos3\\theta=\\cos\\left[90-\\theta+34^{\\circ}\\right]$<\/p>\n $\\Rightarrow$ \u00a0$\\cos3\\theta=\\cos\\left[124^{^{\\circ}}-\\theta\\right]$<\/p>\n $\\Rightarrow$ \u00a0$3\\theta=124^{^{\\circ}}-\\theta$<\/p>\n $\\Rightarrow$ \u00a0$4\\theta=124^{^{\\circ}}$<\/p>\n $\\Rightarrow$ \u00a0$\\theta=31^{^{\\circ}}$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 15:\u00a0<\/b>The value of $\\sin60^{\\circ}\\cos30^{\\circ}-\\cos60^{\\circ}\\sin30^{\\circ}$ is:<\/p>\n a)\u00a01<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a0$\\frac{\\sqrt 3}{2}$<\/p>\n d)\u00a0$\\frac{1}{\\sqrt 2}$<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin60^{\\circ}\\cos30^{\\circ}-\\cos60^{\\circ}\\sin30^{\\circ}=\\frac{\\sqrt{3}}{2}\\times\\frac{\\sqrt{3}}{2}-\\frac{1}{2}\\times\\frac{1}{2}$<\/p>\n $=\\frac{3}{4}-\\frac{1}{4}$<\/p>\n $=\\frac{2}{4}$<\/p>\n $=\\frac{1}{2}$<\/p>\n Hence, the correct answer is Option B<\/p>\n