Enroll to XAT 2023 Mocks & Sectionals <\/a><\/p>\n Question 1:\u00a0<\/b>The value of $(\\cosec 30^\\circ\u00a0– \\tan 45^\\circ) \\cot 60^\\circ \\tan 30^\\circ$ is:<\/p>\n a)\u00a0$\\frac{1}{3}$<\/p>\n b)\u00a0$1$<\/p>\n c)\u00a0$3$<\/p>\n d)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(\\cosec 30^\\circ\u00a0– \\tan 45^\\circ) \\cot 60^\\circ \\tan 30^\\circ$<\/p>\n On put the value,<\/p>\n = $(2 – 1)\\frac{1}{\\sqrt3} \\times \\frac{1}{\\sqrt3}$<\/p>\n = 1\/3<\/p>\n Question 2:\u00a0<\/b>If $7 \\sin^2 \\theta – \\cos^2 \\theta + 2 \\sin \\theta = 2, 0^\\circ < \\theta < 90^\\circ$, then the value of $\\frac{\\sec 2\\theta + \\cot 2\\theta}{\\cosec 2\\theta + \\tan 2\\theta}$ is:<\/p>\n a)\u00a0$\\frac{2\\sqrt{3} + 1}{3}$<\/p>\n b)\u00a01<\/p>\n c)\u00a0$\\frac{1}{5}(1 + 2\\sqrt{3})$<\/p>\n d)\u00a0$\\frac{2}{5}(1 + \\sqrt{3})$<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $7 \\sin^2 \\theta – \\cos^2 \\theta + 2 \\sin \\theta = 2$<\/p>\n $\\Rightarrow$\u00a0$7 \\sin^{2} \\theta – \\cos^{2} \\theta + 2 \\sin \\theta – 2 = 0$<\/p>\n $\\Rightarrow$ $7 \\sin^{2} \\theta – (1 – \\sin^{2} \\theta) + 2 \\sin \\theta – 2 = 0$<\/p>\n $\\Rightarrow$ $8 \\sin^{2} \\theta + 2 \\sin \\theta – 3 = 0$<\/p>\n $\\Rightarrow$ $8 \\sin^{2} \\theta + 6 \\sin \\theta – 4\\sin \\theta- 3 = 0$<\/p>\n $\\Rightarrow$ $2\\sin \\theta(4 \\sin\\theta + 3) -1(4 \\sin\\theta + 3) = 0$<\/p>\n $\\Rightarrow$ $(2\\sin \\theta – 1)(4 \\sin\\theta + 3) = 0$<\/p>\n For $ 0^\\circ < \\theta < 90^\\circ$,<\/p>\n $\\sin \\theta = 1\/2$<\/p>\n $\\theta = 30\\degree$<\/p>\n $\\frac{\\sec 2 \\times 30+ \\cot 2\\times 30}{\\cosec 2\\times 30 + \\tan 2\\times 30}$<\/p>\n $\\Rightarrow$ $\\frac{\\sec 60\u00a0+ \\cot 60}{\\cosec 60\u00a0+ \\tan 60}$ $\\Rightarrow$ $\\frac{2\\sqrt3 + 1}{2+ 3}$<\/p>\n $\\Rightarrow$\u00a0$\\frac{2\\sqrt3 + 1}{5}$<\/p>\n Question 3:\u00a0<\/b>The expression $3 \\sec^2 \\theta \\tan^2 \\theta + \\tan^6 \\theta – \\sec^6 \\theta$ is equal to:<\/p>\n a)\u00a0-2<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a0-1<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $3 \\sec^2 \\theta \\tan^2 \\theta + \\tan^6 \\theta – \\sec^6 \\theta$<\/p>\n = $-3 \\sec^2 \\theta \\tan^2 \\theta(\\tan^2 \\theta –\u00a0\\sec^2 \\theta) + \\tan^6 \\theta – \\sec^6 \\theta$<\/p>\n ($\\because \\tan^2 \\theta – \\sec^2 \\theta = -1$)<\/p>\n =\u00a0($\\tan^2 \\theta – \\sec^2 \\theta$)^3<\/p>\n ($\\because(a – b)^3 = a^3\u00a0– b^3 – 3ab(a – b)$)<\/p>\n = $(-1)^3$ = -1<\/p>\n Question 4:\u00a0<\/b>The value of $\\frac{\\tan^2 \\theta – \\sin^2 \\theta}{2 + \\tan^2 \\theta + \\cot^2 \\theta}$ is:<\/p>\n a)\u00a0$\\cosec^6 \\theta$<\/p>\n b)\u00a0$\\cos^4 \\theta$<\/p>\n c)\u00a0$\\sin^6 \\theta$<\/p>\n d)\u00a0$\\sec^4 \\theta$<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\tan^2 \\theta – \\sin^2 \\theta}{2 + \\tan^2 \\theta + \\cot^2 \\theta}$<\/p>\n = $\\frac{\\tan^2 \\theta – \\sin^2 \\theta}{1 + \\tan^2 \\theta + 1 +\u00a0\\cot^2 \\theta}$<\/p>\n = $\\frac{\\tan^2 \\theta – \\sin^2 \\theta}{\\sec^2 \\theta + \\cosec^2 \\theta}$<\/p>\n = $\\frac{\\tan^2 \\theta – \\sin^2 \\theta}{\\frac{1}{\\cos^2 \\theta} +\\frac{1}{\\sin^2 \\theta}}$<\/p>\n = $\\frac{(\\tan^2 \\theta – \\sin^2 \\theta)(\\sin^2\\theta\\cos^2 \\theta)}{\\cos^2 \\theta + \\sin^2 \\theta}$<\/p>\n = $(\\tan^2 \\theta – \\sin^2 \\theta)(\\sin^2\\theta\\cos^2 \\theta)$<\/p>\n = $ \\sin^4 \\theta –\u00a0\\sin^4 \\theta\\cos^2 \\theta$<\/p>\n = $ \\sin^4 \\theta(1 – \\cos^2 \\theta)$ = $ \\sin^4 \\theta \\sin^2 \\theta =\u00a0 \\sin^6 \\theta$<\/p>\n Question 5:\u00a0<\/b>The value of $\\frac{1 – 2 \\sin^2 \\theta \\cos^2 \\theta}{\\sin^4 \\theta + \\cos^4 \\theta} – 1$ is:<\/p>\n a)\u00a0-1<\/p>\n b)\u00a01<\/p>\n c)\u00a0$-2 \\sin^2 \\theta \\cos^2 \\theta$<\/p>\n d)\u00a00<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{1 – 2 \\sin^2 \\theta \\cos^2 \\theta}{\\sin^4 \\theta + \\cos^4 \\theta} – 1$<\/p>\n =\u00a0$\\frac{1 – 2 \\sin^2 \\theta \\cos^2 \\theta –\u00a0\\sin^4 \\theta – \\cos^4 \\theta}{\\sin^4 \\theta + \\cos^4 \\theta}$<\/p>\n = $\\frac{1 – (sin^2 \\theta + \\cos^2 \\theta)^2 }{\\sin^4 \\theta + \\cos^4 \\theta}$<\/p>\n = $\\frac{1 – 1}{\\sin^4 \\theta + \\cos^4 \\theta}$ = 0<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>What is the value of $\\sin 30^\\circ + \\cos 30^\\circ – \\tan 45^\\circ $?<\/p>\n a)\u00a0$\\frac{\\sqrt{3} – 1}{2}$<\/p>\n b)\u00a0$\\frac{\\sqrt{2} + 1}{\\sqrt{2}}$<\/p>\n c)\u00a0$\\frac{\\sqrt{3} + 1}{2}$<\/p>\n d)\u00a0$\\frac{1 – \\sqrt{3}}{2}$<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sin 30^\\circ + \\cos 30^\\circ – \\tan 45^\\circ $<\/p>\n = $1\/2 + \\frac{\\sqrt3}{2} – 1$<\/p>\n = $\\frac{1 + \\sqrt3 – 2}{2}$<\/p>\n = $\\frac{\\sqrt3 – 1}{2}$<\/p>\n Question 7:\u00a0<\/b>In the given figure, $\\cos\\theta$ is equal to: a)\u00a0$\\frac{5}{13}$<\/p>\n b)\u00a0$\\frac{12}{13}$<\/p>\n c)\u00a0$\\frac{5}{12}$<\/p>\n d)\u00a0$\\frac{12}{5}$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n In\u00a0$\\triangle$PQR,<\/p>\n QR$^2$ + PR$^2$ = PQ$^2$<\/p>\n $\\Rightarrow$\u00a0 12$^2$ + PR$^2$ = 13$^2$<\/p>\n $\\Rightarrow$\u00a0 144 + PR$^2$ = 169<\/p>\n $\\Rightarrow$\u00a0 PR$^2$ = 25<\/p>\n $\\Rightarrow$\u00a0 PR = 5 units<\/p>\n $\\therefore\\ $\u00a0$\\cos\\theta=\\frac{\\text{adjacent side}}{\\text{hypotenuse}}=\\frac{\\text{PR}}{\\text{PQ}}=\\frac{5}{13}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 8:\u00a0<\/b>Solve the following. a)\u00a02<\/p>\n b)\u00a0-2<\/p>\n c)\u00a0-1<\/p>\n d)\u00a01<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\sin 40^\\circ}{\\cos 50^\\circ} + \\frac{\\cosec 50^\\circ}{\\sec 40^\\circ} – 4 \\cos 50^\\circ \\cosec 40^\\circ$<\/p>\n =$\\frac{\\sin 40^\\circ}{\\cos(90^\\circ –\u00a040^\\circ)} + \\frac{\\cosec(90^\\circ – 40^\\circ)}{\\sec 40^\\circ} – 4 \\cos(90^\\circ – 40^\\circ) \\cosec 40^\\circ$<\/p>\n =$\\frac{\\sin 40^\\circ}{\\sin 40^\\circ} + \\frac{\\sec 40^\\circ}{\\sec 40^\\circ} – 4 \\sin40^\\circ \\cosec 40^\\circ$<\/p>\n = 1 +\u00a01 – 4 = -2<\/p>\n Question 9:\u00a0<\/b>If $x \\cos A – y \\sin A = 1$ and $x \\sin A + y \\cos A = 4$, then the value of\u00a0$17x^{2} + 17y^{2}$ is:<\/p>\n a)\u00a00<\/p>\n b)\u00a07<\/p>\n c)\u00a049<\/p>\n d)\u00a0289<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x \\cos A – y \\sin A=1$<\/p>\n Take the square of both sides,<\/p>\n $(x \\cos A – y \\sin A)^2=1$<\/p>\n $(x \\cos A)^2 – 2xy \\sin A \\cos A +\u00a0(y \\sin A)^2\u00a0= 1$ —(1)<\/p>\n ($\\because (a + b)^2 = a^2 + b^2 + 2ab$)<\/p>\n $x \\sin A + y \\cos A = 4$<\/p>\n Take the square of both sides,<\/p>\n $(x \\sin A + y \\cos A)^2=16$<\/p>\n $(x \\sin A)^2 + 2xy \\sin A \\cos A + (y \\cos A)^2 =16$ —(2)<\/p>\n Addition of eq(1) and eq(2),<\/p>\n $(x \\cos A)^2 +\u00a0(x \\sin A)^2 +\u00a0(y \\sin A)^2 +\u00a0(y \\cos A)^2 = 17$<\/p>\n $x^2 + y^2 = 17$<\/p>\n $17x^{2}+17y^{2} = 17 \\times 17$<\/p>\n $17x^{2}+17y^{2} = 289$<\/p>\n Question 10:\u00a0<\/b>If $(2 \\sin A + \\cosec A) = 2 \\sqrt{2}$, $0^\\circ < A < 90^\\circ$ then the value of\u00a0$2(\\sin^{4}A + \\cos^{4}A)$ is:<\/p>\n a)\u00a02<\/p>\n b)\u00a01<\/p>\n c)\u00a04<\/p>\n d)\u00a00<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(2 \\sin A + \\cosec A) = 2 \\sqrt{2}$<\/p>\n To find the value A, we satisfy the above equation so put the value of A = 45$\\degree$<\/p>\n $(2 \\sin 45 \\degree\u00a0+ \\cosec 45 \\degree) = 2 \\sqrt{2}$<\/p>\n $(2 \\times \\frac{1}{\\sqrt{2}} + \\sqrt{2}) = 2 \\sqrt{2}$<\/p>\n $ 2\\sqrt{2} =\u00a02\\sqrt{2}$<\/p>\n $2(\\sin^{4}A + \\cos^{4}A)$<\/p>\n =\u00a0$2(\\sin^{4}45 \\degree + \\cos^{4}45\\degree)$<\/p>\n =\u00a0$2((\\frac{1}{\\sqrt{2}})^{4} + (\\frac{1}{\\sqrt{2}})^{4})$<\/p>\n = $2(\\frac{1}{4} + \\frac{1}{4}) = 2(\\frac{1}{2}) = 1$<\/p>\n
\n$\\Rightarrow$ $\\frac{2+ \\frac{1}{\\sqrt3}}{\\frac{2}{\\sqrt3} + \\sqrt3}$<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Q13_9rilqPv.png\")
<\/p>\n
\n$\\frac{\\sin 40^\\circ}{\\cos 50^\\circ} + \\frac{\\cosec 50^\\circ}{\\sec 40^\\circ} – 4 \\cos 50^\\circ \\cosec 40^\\circ$<\/p>\n