Download Trignometry Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $\\frac{\\sin\\theta+\\cos\\theta}{\\sin\\theta-\\cos\\theta}=5$, then the value of $\\frac{4\\sin^2\\theta+3}{2\\cos^2\\theta+2}$ is:<\/p>\n a)\u00a0$\\frac{75}{17}$<\/p>\n b)\u00a0$\\frac{75}{34}$<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$\\frac{3}{2}$<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\sin\\theta+\\cos\\theta}{\\sin\\theta-\\cos\\theta}=5$<\/p>\n $\\sin\\theta+\\cos\\theta=5\\sin\\theta\\ -5\\cos\\theta\\ $<\/p>\n $4\\sin\\theta=6\\cos\\theta\\ $<\/p>\n $\\tan\\theta\\ =\\frac{3}{2}$<\/p>\n $\\sec\\theta\\ =\\sqrt{\\left(\\frac{3}{2}\\right)^2+1}=\\frac{\\sqrt{13}}{2}$<\/p>\n $\\cos\\theta\\ =\\frac{2}{\\sqrt{13}}$<\/p>\n $\\sin\\theta\\ =\\sqrt{1-\\left(\\frac{2}{\\sqrt{13}}\\right)^2}=\\frac{3}{\\sqrt{13}}$<\/p>\n $\\frac{4\\sin^2\\theta+3}{2\\cos^2\\theta+2}=\\frac{4\\left(\\frac{3}{\\sqrt{13}}\\right)^2+3}{2\\left(\\frac{2}{\\sqrt{13}}\\right)^2+2}$<\/p>\n $=\\frac{\\frac{36}{13}+3}{\\frac{8}{13}+2}$<\/p>\n $=\\frac{\\frac{36+39}{13}}{\\frac{8+26}{13}}$<\/p>\n $=\\frac{75}{34}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 2:\u00a0<\/b>Find the value of $\\frac{\\tan^2 30^\\circ}{\\sec^2 30^\\circ} + \\frac{\\cosec^2 45^\\circ}{\\cot^2 45^\\circ} – \\frac{\\sec^2 60^\\circ}{\\cosec^2 60^\\circ}$<\/p>\n a)\u00a0$-\\frac{3}{4}$<\/p>\n b)\u00a0$\\frac{5}{4}$<\/p>\n c)\u00a0$\\frac{13}{4}$<\/p>\n d)\u00a0$\\frac{23}{12}$<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\tan^230^{\\circ}}{\\sec^230^{\\circ}}+\\frac{\\operatorname{cosec}^245^{\\circ}}{\\cot^245^{\\circ}}-\\frac{\\sec^260^{\\circ}}{\\operatorname{cosec}^260^{\\circ}}=\\frac{\\left(\\frac{1}{\\sqrt{3}}\\right)^2}{\\left(\\frac{2}{\\sqrt{3}}\\right)^2}+\\frac{\\left(\\sqrt{2}\\right)^2}{\\left(1\\right)^2}-\\frac{\\left(2\\right)^2}{\\left(\\frac{2}{\\sqrt{3}}\\right)^2}$<\/p>\n $=\\frac{\\frac{1}{3}}{\\frac{4}{3}}+\\frac{2}{1}-\\frac{4}{\\frac{4}{3}}$<\/p>\n $=\\frac{1}{4}+2-\\frac{4\\times3}{4}$<\/p>\n $=\\frac{-3}{4}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>If $\\sin^2 \\theta – \\cos^2 \\theta – 3 \\sin \\theta + 2 = 0, 0^\\circ < \\theta < 90^\\circ$, then what is the value of $\\frac{1}{\\sqrt{\\sec \\theta – \\tan \\theta}}$ is:<\/p>\n a)\u00a0$\\sqrt[4]{3}$<\/p>\n b)\u00a0$\\sqrt[2]{2}$<\/p>\n c)\u00a0$\\sqrt[2]{3}$<\/p>\n d)\u00a0$\\sqrt[4]{2}$<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin^2\\theta-\\cos^2\\theta-3\\sin\\theta+2=0$<\/p>\n $\\sin^2\\theta-\\left(1-\\sin^2\\theta\\ \\right)-3\\sin\\theta+2=0$<\/p>\n $2\\sin^2\\theta-3\\sin\\theta+1=0$<\/p>\n $2\\sin^2\\theta-2\\sin\\theta-\\sin\\theta\\ +1=0$<\/p>\n $2\\sin\\theta\\ \\left(\\sin\\theta\\ -1\\right)-1\\left(\\sin\\theta\\ -1\\right)=0$<\/p>\n $\\left(\\sin\\theta\\ -1\\right)\\left(2\\sin\\theta\\ -1\\right)=0$<\/p>\n $\\sin\\theta\\ -1=0$\u00a0 or\u00a0\u00a0$2\\sin\\theta\\ -1=0$<\/p>\n $\\sin\\theta\\ =1$\u00a0 or\u00a0\u00a0$\\sin\\theta\\ =\\frac{1}{2}$<\/p>\n $\\theta\\ =90^{\\circ\\ }$\u00a0 or\u00a0\u00a0$\\theta\\ =30^{\\circ\\ }$<\/p>\n Given, $0^\\circ < \\theta < 90^\\circ$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\theta\\ =30^{\\circ\\ }$<\/p>\n $\\frac{1}{\\sqrt{\\sec\\theta-\\tan\\theta}}=\\frac{1}{\\sqrt{\\sec30^{\\circ\\ }-\\tan30^{\\circ\\ }}}$<\/p>\n $=\\frac{1}{\\sqrt{\\frac{2}{\\sqrt{3}}\\ -\\frac{1}{\\sqrt{3}}}}$<\/p>\n $=\\frac{1}{\\sqrt{\\frac{1}{\\sqrt{3}}}}$<\/p>\n $=\\sqrt[\\ 4]{3}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 4:\u00a0<\/b>If $3 \\sec \\theta + 4 \\cos \\theta – 4\\sqrt{3} = 0$ where $\\theta$ is an acute angle then the value of $\\theta$ is:<\/p>\n a)\u00a0$20^\\circ$<\/p>\n b)\u00a0$30^\\circ$<\/p>\n c)\u00a0$60^\\circ$<\/p>\n d)\u00a0$45^\\circ$<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $3\\sec\\theta+4\\cos\\theta-4\\sqrt{3}=0$<\/p>\n $\\frac{3}{\\cos\\theta\\ }+4\\cos\\theta-4\\sqrt{3}=0$<\/p>\n $4\\cos^2\\theta-4\\sqrt{3}\\cos\\theta\\ +3=0$<\/p>\n $4\\cos^2\\theta-2\\sqrt{3}\\cos\\theta\\ -2\\sqrt{3}\\cos\\theta+3=0$<\/p>\n $2\\cos\\theta\\ \\left(2\\cos\\theta-\\sqrt{3}\\right)\\ -\\sqrt{3}\\left(2\\cos\\theta-\\sqrt{3}\\right)=0$<\/p>\n $\\ \\left(2\\cos\\theta-\\sqrt{3}\\right)\\left(2\\cos\\theta-\\sqrt{3}\\right)=0$<\/p>\n $\\ \\left(2\\cos\\theta-\\sqrt{3}\\right)^2=0$<\/p>\n $\\ 2\\cos\\theta-\\sqrt{3}=0$<\/p>\n $\\cos\\theta=\\frac{\\sqrt{3}}{2}$<\/p>\n $\\theta=30^{\\circ\\ }$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>If $3 \\tan \\theta = 2\\sqrt{3} \\sin \\theta, 0^\\circ < \\theta < 90^\\circ$, then find the value of $2 \\sin^2 2\\theta – 3 \\cos^2 3\\theta$.<\/p>\n a)\u00a01<\/p>\n b)\u00a0$\\frac{3}{2}$<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$-\\frac{3}{2}$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $3\\tan\\theta=2\\sqrt{3}\\sin\\theta$<\/p>\n $3\\frac{\\sin\\theta\\ }{\\cos\\theta\\ }=2\\sqrt{3}\\sin\\theta$<\/p>\n $\\cos\\theta\\ =\\frac{3}{2\\sqrt{3}}$<\/p>\n $\\cos\\theta\\ =\\frac{\\sqrt{3}}{2}$<\/p>\n $\\theta\\ =30^{\\circ\\ }$ [$0^\\circ < \\theta < 90^\\circ$]<\/p>\n $2\\sin^22\\theta-3\\cos^23\\theta=2\\sin^260^{\\circ\\ }-3\\cos^290^{\\circ\\ }$<\/p>\n =\u00a0$2\\left(\\frac{\\sqrt{3}}{2}\\right)^2-3\\left(0\\right)^2$<\/p>\n =\u00a0$\\frac{3}{2}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Enroll to SNAP & NMAT 2022 Crash Course<\/a><\/p>\n Question 6:\u00a0<\/b>Find the value of $\\sin^2 60^\\circ + \\cos^2 30^\\circ – \\sin^2 45^\\circ – 3 \\sin^2 90^\\circ$.<\/p>\n a)\u00a0$\\frac{1}{3}$<\/p>\n b)\u00a0$-1\\frac{3}{4}$<\/p>\n c)\u00a0$-2\\frac{1}{2}$<\/p>\n d)\u00a0$-2$<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin^260^{\\circ}+\\cos^230^{\\circ}-\\sin^245^{\\circ}-3\\sin^290^{\\circ}=\\left(\\frac{\\sqrt{3}}{2}\\right)^2+\\left(\\frac{\\sqrt{3}}{2}\\right)^2-\\left(\\frac{1}{\\sqrt{2}}\\right)^2-3\\left(1\\right)^2$<\/p>\n $=\\frac{3}{4}+\\frac{3}{4}-\\frac{1}{2}-3$<\/p>\n $=\\frac{3+3-2-12}{4}$<\/p>\n $=\\frac{-8}{4}$<\/p>\n $=-2$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>The value of $\\frac{\\sec^2 60^\\circ \\cos^2 45^\\circ + \\cosec^2 30^\\circ}{\\cot 30^\\circ \\sec^2 45^\\circ – \\cosec^2 30^\\circ \\tan 45^\\circ}$ is:<\/p>\n a)\u00a0$-3(2 + \\sqrt{3})$<\/p>\n b)\u00a0$3(2 – \\sqrt{3})$<\/p>\n c)\u00a0$-3(2 – \\sqrt{3})$<\/p>\n d)\u00a0$3(2 + \\sqrt{3})$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\sec^260^{\\circ}\\cos^245^{\\circ}+\\operatorname{cosec}^230^{\\circ}}{\\cot30^{\\circ}\\sec^245^{\\circ}-\\operatorname{cosec}^230^{\\circ}\\tan45^{\\circ}}=\\frac{\\left(2\\right)^2.\\left(\\frac{1}{\\sqrt{2}}\\right)^2+\\left(2\\right)^2}{\\left(\\sqrt{3}\\right)\\left(\\sqrt{2}\\right)^2-\\left(2\\right)^2.\\left(1\\right)}$<\/p>\n $=\\frac{4\\times\\frac{1}{2}+4}{2\\sqrt{3}-4}$<\/p>\n $=\\frac{6}{2\\sqrt{3}-4}$<\/p>\n $=\\frac{3}{\\sqrt{3}-2}$<\/p>\n $=\\frac{3}{\\sqrt{3}-2}\\times\\frac{\\sqrt{3}+2}{\\sqrt{3}+2}$<\/p>\n $=\\frac{3\\left(\\sqrt{3}+2\\right)}{3-4}$<\/p>\n $=-3\\left(2+\\sqrt{3}\\right)$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 8:\u00a0<\/b>If $\\sin^2 \\theta = 2 \\sin \\theta – 1, 0^\\circ \\leq \\theta \\leq 90^\\circ $, then find the value of: $\\frac{1 + \\cosec \\theta}{1 – \\cos \\theta}$.<\/p>\n a)\u00a0-2<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a0-1<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin^2\\theta=2\\sin\\theta-1$<\/p>\n $\\sin^2\\theta-2\\sin\\theta+1=0$<\/p>\n $\\left(\\sin\\theta-1\\right)^2=0$<\/p>\n $\\sin\\theta-1=0$<\/p>\n $\\sin\\theta=1$<\/p>\n $0^{\\circ}\\le\\theta\\le90^{\\circ}$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\theta=90^{\\circ}$<\/p>\n $\\frac{1+\\operatorname{cosec}\\theta}{1-\\cos\\theta}=\\frac{1+\\operatorname{cosec}90^{\\circ\\ }}{1-\\cos90^{\\circ\\ }}$<\/p>\n $=\\frac{1+1\\ }{1-0\\ }$<\/p>\n $=2$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 9:\u00a0<\/b>Simplify $\\sec^2 \\alpha \\left(1 + \\frac{1}{\\cosec \\alpha}\\right)\\left(1 – \\frac{1}{\\cosec \\alpha}\\right)$.<\/p>\n a)\u00a0$\\tan^4 \\alpha$<\/p>\n b)\u00a0$\\sin^2 \\alpha$<\/p>\n c)\u00a01<\/p>\n d)\u00a0-1<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sec^2\\alpha\\left(1+\\frac{1}{\\operatorname{cosec}\\alpha}\\right)\\left(1-\\frac{1}{\\operatorname{cosec}\\alpha}\\right)=\\frac{1}{\\cos^2\\alpha}\\left(1+\\sin\\alpha\\right)\\left(1-\\sin\\alpha\\right)$<\/p>\n $=\\frac{1}{\\cos^2\\alpha}\\left(1-\\sin^2\\alpha\\right)$<\/p>\n $=\\frac{1}{\\cos^2\\alpha}\\left(\\cos^2\\alpha\\right)$<\/p>\n $=1$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 10:\u00a0<\/b>In $\\triangle$ABC, right angled at B, if cot A = $\\frac{1}{2}$, then the value of $\\frac{\\sin A(\\cos C + \\cos A)}{\\cos C(\\sin C – \\sin A)}$ is<\/p>\n a)\u00a03<\/p>\n b)\u00a0-3<\/p>\n c)\u00a0-2<\/p>\n d)\u00a02<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n cot A =\u00a0$\\frac{\\text{Adjacent side}}{\\text{Opposite side}}$\u00a0$\\frac{1}{2}$<\/p>\n $\\frac{\\sin A(\\cos C + \\cos A)}{\\cos C(\\sin C – \\sin A)}$ =\u00a0$\\frac{\\frac{2}{\\sqrt{5}}\\left(\\frac{2}{\\sqrt{5}}+\\frac{1}{\\sqrt{5}}\\right)}{\\frac{2}{\\sqrt{5}}\\left(\\frac{1}{\\sqrt{5}}-\\frac{2}{\\sqrt{5}}\\right)}$<\/p>\n =\u00a0$\\frac{\\left(\\frac{3}{\\sqrt{5}}\\right)}{\\left(-\\frac{1}{\\sqrt{5}}\\right)}$<\/p>\n = -3<\/p>\n Hence, the correct answer is Option B<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 11:\u00a0<\/b>If $\\tan \\theta + 3 \\cot \\theta – 2\\sqrt{3} = 0, 0^\\circ < \\theta < 90^\\circ$, then what is the value of $(\\cosec^2 \\theta + \\cos^2 \\theta)?$<\/p>\n a)\u00a0$\\frac{2}{3}$<\/p>\n b)\u00a0$\\frac{19}{12}$<\/p>\n c)\u00a0$\\frac{14}{3}$<\/p>\n d)\u00a0$\\frac{11}{12}$<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\tan\\theta+3\\cot\\theta-2\\sqrt{3}=0$<\/p>\n $\\tan\\theta+\\frac{3}{\\tan\\theta\\ }-2\\sqrt{3}=0$<\/p>\n $tan^2\\theta-2\\sqrt{3}\\tan\\theta\\ +3=0$<\/p>\n $\\left(\\tan\\theta\\ -\\sqrt{3}\\right)^2=0$<\/p>\n $\\tan\\theta\\ -\\sqrt{3}=0$<\/p>\n $\\tan\\theta\\ =\\sqrt{3}$<\/p>\n $0^\\circ < \\theta < 90^\\circ$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\theta\\ =60^{\\circ\\ }$<\/p>\n $\\operatorname{cosec}^2\\theta+\\cos^2\\theta=\\operatorname{cosec}^260^{\\circ\\ }+\\cos^260^{\\circ\\ }$<\/p>\n =\u00a0$\\left(\\frac{2}{\\sqrt{3}}\\right)^2\\ +\\left(\\frac{1}{2}\\right)^2$<\/p>\n =\u00a0$\\frac{4}{3}\\ +\\frac{1}{4}$<\/p>\n =\u00a0$\\frac{16+3}{12}$<\/p>\n =\u00a0$\\frac{19}{12}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 12:\u00a0<\/b>If $\\sin \\alpha + \\sin \\beta = \\cos \\alpha + \\cos \\beta = 1$, then $\\sin \\alpha + \\cos \\alpha =$?<\/p>\n a)\u00a0-1<\/p>\n b)\u00a00<\/p>\n c)\u00a01<\/p>\n d)\u00a02<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin\\alpha+\\sin\\beta=1$<\/p>\n $\\sin^2\\alpha+\\sin^2\\beta+2\\sin\\alpha\\ \\sin\\beta\\ =1$……(1)<\/p>\n $\\cos\\alpha+\\cos\\beta=1$<\/p>\n $\\cos^2\\alpha+\\cos^2\\beta+2\\cos\\alpha\\ \\cos\\beta\\ =1$……(2)<\/p>\n Adding (1) and (2),<\/p>\n $\\left(\\sin^2\\alpha\\ +\\cos^2\\alpha\\right)+\\left(\\sin^2\\beta\\ +\\cos^2\\beta\\right)+2\\sin\\alpha\\ \\sin\\beta\\ +2\\cos\\alpha\\ \\cos\\beta\\ =1+1$<\/p>\n $1+1+2\\sin\\alpha\\ \\sin\\beta\\ +2\\cos\\alpha\\ \\cos\\beta\\ =2$<\/p>\n $2\\left[\\cos\\alpha\\ \\cos\\beta+\\sin\\alpha\\ \\sin\\beta\\right]=0$<\/p>\n $\\cos\\left(\\beta-\\alpha\\right)=0$<\/p>\n $\\beta-\\alpha=90^{\\circ\\ }$<\/p>\n $\\beta\\ =90^{\\circ\\ }+\\alpha\\ $<\/p>\n $\\sin\\alpha+\\sin\\beta=1$<\/p>\n $\\sin\\alpha+\\sin\\left(90^{\\circ}-\\alpha\\ \\right)=1$<\/p>\n $\\sin\\alpha+\\cos\\alpha=1$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 13:\u00a0<\/b>Find the value of $\\operatorname{cosec}(60^{\\circ}+A)-\\sec(30^{\\circ}-A)+\\frac{\\operatorname{cosec}49^{\\circ}}{\\sec41^{\\circ}}$.<\/p>\n a)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a0-1<\/p>\n d)\u00a02<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\operatorname{cosec}(60^{\\circ}+A)-\\sec(30^{\\circ}-A)+\\frac{\\operatorname{cosec}49^{\\circ}}{\\sec41^{\\circ}}$<\/p>\n =\u00a0$\\operatorname{cosec}(60^{\\circ}+A)-\\sec(90^{\\circ}-60^{\\circ}-A)+\\frac{\\operatorname{cosec}49^{\\circ}}{\\sec\\left(90-49\\right)^{\\circ}}$<\/p>\n $\\left[\\sec\\left(90\\ -\\theta\\right)=\\operatorname{cosec}\\theta\\right]$<\/p>\n =\u00a0$\\operatorname{cosec}\\left(60^{\\circ}+A\\right)-\\sec\\left(90^{\\circ}-\\left(60^{\\circ}+A\\right)\\right)+\\frac{\\operatorname{cosec}49^{\\circ}}{\\operatorname{cosec}49^{\\circ}}$<\/p>\n =\u00a0$\\operatorname{cosec}\\left(60^{\\circ}+A\\right)-\\operatorname{cosec}\\left(60^{\\circ}+A\\right)+1$<\/p>\n = 1<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 14:\u00a0<\/b>If $\\frac{1}{1 – \\sin \\theta} + \\frac{1}{1 + \\sin \\theta} = 4 \\sec \\theta, 0^\\circ < \\theta < 90^\\circ$, then the value of $\\cot\\theta+\\operatorname{cosec}\\theta$ is:<\/p>\n a)\u00a0$\\frac{4\\sqrt{3}}{3}$<\/p>\n b)\u00a0$\\sqrt{3}$<\/p>\n c)\u00a0$\\frac{5\\sqrt{3}}{3}$<\/p>\n d)\u00a0$3\\sqrt{3}$<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{1}{1-\\sin\\theta}+\\frac{1}{1+\\sin\\theta}=4\\sec\\theta$<\/p>\n $\\frac{1+\\sin\\theta\\ +1-\\sin\\theta\\ }{1-\\sin^2\\theta}=\\frac{4}{\\cos\\theta}$<\/p>\n $\\frac{2}{\\cos^2\\theta}=\\frac{4}{\\cos\\theta}$<\/p>\n $\\cos\\theta=\\frac{1}{2}$<\/p>\n $0^\\circ < \\theta < 90^\\circ$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\theta=60^{\\circ}$<\/p>\n $\\cot\\theta+\\operatorname{cosec}\\theta=\\cot60^{\\circ}+\\operatorname{cosec}60^{\\circ}$<\/p>\n =\u00a0$\\frac{1}{\\sqrt{3}}+\\frac{2}{\\sqrt{3}}$<\/p>\n =\u00a0$\\frac{3}{\\sqrt{3}}$<\/p>\n =\u00a0$\\sqrt{3}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 15:\u00a0<\/b>$(\\operatorname{cosec}A-\\cot A)(1+\\cos A)=?$<\/p>\n a)\u00a0$\\cos A$<\/p>\n b)\u00a0$\\sin A$<\/p>\n c)\u00a0$\\cot A$<\/p>\n d)\u00a0$\\cosec A$<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(\\operatorname{cosec}A-\\cot A)(1+\\cos A)=\\left(\\frac{1}{\\sin A}-\\frac{\\cos A}{\\sin A}\\right)\\left(1+\\cos A\\right)$<\/p>\n =\u00a0$\\left(\\frac{1-\\cos A}{\\sin A}\\right)\\left(1+\\cos A\\right)$<\/p>\n =\u00a0$\\left(\\frac{1-\\cos^2A}{\\sin A}\\right)$<\/p>\n =\u00a0$\\frac{\\sin^2A}{\\sin A}$<\/p>\n =\u00a0$\\sin A$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 16:\u00a0<\/b>If $0^\\circ < \\theta < 90^\\circ, \\sqrt{\\frac{\\sec^2 \\theta + \\cosec^2 \\theta}{\\tan^2 \\theta – \\sin^2 \\theta}}$ is equal to:<\/p>\n a)\u00a0$\\sec^3 \\theta$<\/p>\n b)\u00a0$\\cosec^3 \\theta$<\/p>\n c)\u00a0$\\sin^2 \\theta$<\/p>\n d)\u00a0$\\sec^2 \\theta$<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sqrt{\\frac{\\sec^2\\theta+\\operatorname{cosec}^2\\theta}{\\tan^2\\theta-\\sin^2\\theta}}=\\sqrt{\\frac{\\frac{1}{\\cos^2\\theta}+\\frac{1}{\\sin^2\\theta}}{\\frac{\\sin^2\\theta\\ }{\\cos^2\\theta}-\\sin^2\\theta}}$<\/p>\n $=\\sqrt{\\frac{\\frac{\\sin^2\\theta+\\cos^2\\theta\\ }{\\cos^2\\theta\\sin^2\\theta}}{\\frac{\\sin^2\\theta-\\sin^2\\theta\\cos^2\\theta\\ }{\\cos^2\\theta}}}$<\/p>\n $=\\sqrt{\\frac{1\\ }{\\cos^2\\theta\\sin^2\\theta}\\times\\frac{\\cos^2\\theta\\ }{\\sin^2\\theta\\left(1-\\cos^2\\theta\\right)\\ }}$<\/p>\n $=\\sqrt{\\frac{1\\ }{\\sin^2\\theta}\\times\\frac{1}{\\sin^2\\theta\\left(\\sin^2\\theta\\right)\\ }}$<\/p>\n $=\\sqrt{\\frac{1\\ }{\\sin^6\\theta}\\ }$<\/p>\n $=\\sqrt{\\operatorname{cosec}^6\\theta\\ }$<\/p>\n $=\\operatorname{cosec}^3\\theta\\ $<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 17:\u00a0<\/b>If $\\frac{\\sin^2 \\theta}{\\tan^2 \\theta – \\sin^2 \\theta} = 5, \\theta$ is an acute angle, then the value of $\\frac{24\\sin^2\\theta-15\\sec^2\\theta}{6\\operatorname{cosec}^2\\theta-7\\cot^2\\theta}$ is:<\/p>\n a)\u00a0-2<\/p>\n b)\u00a02<\/p>\n c)\u00a0-14<\/p>\n d)\u00a014<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\sin^2\\theta}{\\tan^2\\theta-\\sin^2\\theta}=5$<\/p>\n $\\frac{\\sin^2\\theta}{\\frac{\\sin^2\\theta\\ }{\\cos^2\\theta\\ }-\\sin^2\\theta}=5$<\/p>\n $\\frac{\\sin^2\\theta}{\\sin^2\\theta\\ \\left(\\frac{1\\ }{\\cos^2\\theta\\ }-1\\right)}=5$<\/p>\n $\\frac{\\cos^2\\theta}{1-\\cos^2\\theta\\ }=5$<\/p>\n $\\frac{\\cos^2\\theta}{\\sin^2\\theta\\ \\ }=5$<\/p>\n $\\cot^2\\theta\\ \\ =5$<\/p>\n $\\operatorname{cosec}^2\\theta\\ =1+\\cot^2\\theta\\ =1+5=6$<\/p>\n $\\sin^2\\theta=\\frac{1}{\\operatorname{cosec}^2\\theta}\\ =\\frac{1}{6}$<\/p>\n $\\cos^2\\theta\\ =1-\\sin^2\\theta=1-\\frac{1}{6}=\\frac{5}{6}$<\/p>\n $\\sec^2\\theta\\ =\\frac{1}{\\cos^2\\theta\\ }=\\frac{6}{5}$<\/p>\n $\\frac{24\\sin^2\\theta-15\\sec^2\\theta}{6\\operatorname{cosec}^2\\theta-7\\cot^2\\theta}=\\frac{24\\left(\\frac{1}{6}\\right)-15\\left(\\frac{6}{5}\\right)}{6\\left(6\\right)-7\\left(5\\right)}$<\/p>\n $=\\frac{4-18}{36-35}$<\/p>\n $=-14$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 18:\u00a0<\/b>The value of $\\sec^4 \\theta (1 – \\sin^4 \\theta) – 2 \\tan^2 \\theta$ is:<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a01<\/p>\n c)\u00a0-1<\/p>\n d)\u00a00<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sec^4\\theta(1-\\sin^4\\theta)-2\\tan^2\\theta=\\sec^4\\theta\\ -\\sec^4\\theta\\ \\sin^4\\theta\\ -2\\tan^2\\theta\\ $<\/p>\n $=\\sec^4\\theta\\ -\\frac{\\sin^4\\theta}{\\cos^4\\theta\\ }\\ -2\\tan^2\\theta\\ $<\/p>\n $=\\sec^4\\theta\\ -\\tan^4\\theta-2\\tan^2\\theta\\ $<\/p>\n $=\\left(\\sec^2\\theta\\ +\\tan^2\\theta\\ \\right)\\left(\\sec^2\\theta\\ -\\tan^2\\theta\\ \\right)-2\\tan^2\\theta\\ $<\/p>\n $=\\left(\\sec^2\\theta\\ +\\tan^2\\theta\\ \\right)\\left(1\\right)-2\\tan^2\\theta\\ $<\/p>\n $=\\sec^2\\theta\\ +\\tan^2\\theta\\ -2\\tan^2\\theta\\ $<\/p>\n $=\\sec^2\\theta\\ -\\tan^2\\theta\\ $<\/p>\n $=1$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 19:\u00a0<\/b>The value of $\\sin^2 60^\\circ \\cos^2 45^\\circ + 2 \\tan^2 60^\\circ – \\cosec^2 30^\\circ$ is equal to:<\/p>\n a)\u00a0$\\frac{17}{24}$<\/p>\n b)\u00a0$\\frac{19}{8}$<\/p>\n c)\u00a0$-\\frac{17}{24}$<\/p>\n d)\u00a0$-\\frac{19}{8}$<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin^260^{\\circ}\\cos^245^{\\circ}+2\\tan^260^{\\circ}-\\operatorname{cosec}^230^{\\circ}=\\left(\\frac{\\sqrt{3}}{2}\\right)^2.\\left(\\frac{1}{\\sqrt{2}}\\right)^2+2\\left(\\sqrt{3}\\right)^2-\\left(2\\right)^2$<\/p>\n $=\\frac{3}{4}.\\frac{1}{2}+2\\left(3\\right)-4$<\/p>\n $=\\frac{3}{8}+6-4$<\/p>\n $=\\frac{3}{8}+2$<\/p>\n $=\\frac{3+16}{8}$<\/p>\n $=\\frac{19}{8}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 20:\u00a0<\/b>The value of $\\frac{\\tan13^{\\circ}\\tan36^{\\circ}\\tan45^{\\circ}\\tan54^{\\circ}\\tan77^{\\circ}}{2\\sec^260^{\\circ}(\\sin^260^{\\circ}-3\\cos60^{\\circ}+2)}$ is:<\/p>\n a)\u00a0$\\frac{1}{10}$<\/p>\n b)\u00a0$-\\frac{1}{4}$<\/p>\n c)\u00a0$\\frac{1}{4}$<\/p>\n d)\u00a0$-\\frac{1}{10}$<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\tan13^{\\circ}\\tan36^{\\circ}\\tan45^{\\circ}\\tan54^{\\circ}\\tan77^{\\circ}}{2\\sec^260^{\\circ}(\\sin^260^{\\circ}-3\\cos60^{\\circ}+2)}=\\frac{\\tan13^{\\circ}\\tan36^{\\circ}\\left(1\\right)\\tan\\left(90-36^{\\circ}\\right)\\tan\\left(90-13^{\\circ\\ }\\right)}{2\\left(2\\right)^2\\left(\\left(\\frac{\\sqrt{3}}{2}\\right)^2-3\\left(\\frac{1}{2}\\right)+2\\right)}$<\/p>\n $=\\frac{\\tan13^{\\circ}\\tan36^{\\circ}\\cot36^{\\circ}\\cot13^{\\circ\\ }}{2\\left(4\\right)\\left(\\frac{3}{4}-\\frac{3}{2}+2\\right)}$<\/p>\n $=\\frac{1}{8\\left(\\frac{3-6+8}{4}\\right)}$<\/p>\n $=\\frac{1}{2\\left(5\\right)}$<\/p>\n $=\\frac{1}{10}$<\/p>\n Hence, the correct answer is Option A<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_tSHdR1E.png\")
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