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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
1. tanx =43 |
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| 2. |
Let Us Practise 7Aind the unknown values in the followingCP-71800, SP 1750, Profit/Loss-?1. |
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| 3. |
Let Us Practise 4.3A farmer has 5,708 mango trees in his garden. He planted 2,856 trees monnumber of trees in the garden now?te planted 2,856 trees more. What is the total |
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Answer» 5708+2856=72564 there are no no of mangos added more n 72564 are the total no. of mangoes |
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| 4. |
Integration e*(sinx + cosx)dx |
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| 5. |
\frac { \operatorname { cos } ( A + B ) } { \operatorname { cos } ( A - B ) } = \frac { 1 - \operatorname { tan } A \cdot \operatorname { tan } B } { 1 + \operatorname { tan } A . \operatorname { tan } B } |
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| 6. |
A Hemispherical Dome Of a Building Needs to be Painted. If Circumference Of the Base Of the dome is 17.6m. Find the Cost of Painting Of it Given Cost of Painting is Rs 4 Per 100cm2 use Pie as 22/7 |
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| 7. |
integration root sinx |
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Answer» sorry this is not a prefect answer try again |
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| 8. |
1 by 1+Sinx integration |
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| 9. |
28.A building is in the form of a cylinder surmounted by a hemispherical dome and contains194121 m2 of air. If the internal diameter of the dome is equal to its total height above thefloor, find the height of the building. |
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| 10. |
A building is in the form of a cylinder surmounted by a hemisphere valted dome andcontains 41 m3 of air. If the internal diameter of dome is equal to its total heightabove the floor, find the height of the building.21 |
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Answer» Please like the solution 👍 ✔️👍 Let r be the radius of hemisphere & Cylinder and h be the height of the Cylinder, H be the height of the Total building.GIVEN :Volume of air = 880/21 m³Internal diameter (d) = HInternal Diameter = 2r = HTotal Height of the building (H) = 2r……(1)Height of the building = height of the cylinder + radius of the hemispherical DomeH = h + r 2r = h +r [from eq 1]2r -r = hr = h ……………..(2)Volume of air inside the building = Volume of cylindrical portion + Volume of hemispherical portionπr²h + (2πr³/3)= 880/21π(h)²h + (2π(h)³/3)= 880/21[From eq 2, r= h]πh³ + ⅔ πh³ = 880/21πh³(1+⅔) = 880/21πh³[(3+2)/3] = 880/21πh³[5/3] = 880/2122/7 × h³ × 5/3 = 880/21h³ = (880 ×3 ×7) / 21 × 22 × 5h³ = 40 /5 = 8h³ = 8h = ³√8 = ³√2×2×2h = 2 mh= r = 2 m [From eq 2, r= h]Total height of the building( H) = 2r = 2×2 = 4 mHence, the Total height of the building is 4m. HOPE THIS WILL HELP YOU… |
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| 11. |
\operatorname { tan } A = \frac { 1 } { 35 } , \operatorname { tan } B = \frac { 18 } { 17 } \text { then } \operatorname { tan } ( A - B ) = |
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| 12. |
1LJ‘ (का कक-1() cosx(tan™ x)+C (b) ~cosx(tan™ x)+C(€) 2xcosx(tan™ x)+C०) ~2xcosx(tan” x)+C |
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| 13. |
\operatorname { tan } A = \frac { 1 } { 2 } , \operatorname { tan } B = \frac { 1 } { 3 } , \text { then } \operatorname { tan } ( A + B ) |
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Answer» tan(A+B)=tanA+tanB/1-tanAtanB=1/2+1/3/1-1/2*1/3=5/6-1=5/5=1 |
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| 14. |
\operatorname { tan } A = \frac { 1 } { 2 } \text { and } \operatorname { tan } B = \frac { 1 } { 3 } , \text { find } \text { value of } \operatorname { tan } ( 2 A + B ) |
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Answer» wrong answer is 3 okok check now... it's 3.. . ok |
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| 15. |
3. cotx=-43 |
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Answer» Cotx=-√3cotx=cot120=cot300 x=120,300 |
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| 16. |
tan–¹= ( cosecx — cotx) |
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| 17. |
(tanx + cotx) dx. |
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Answer» ∫(tanx + cotx)²dx = ∫tan² x + 2tanx cot x + cot² x) dx = ∫(tan² x + cot² x + 2)dx = ∫(tan² x + 1 + cot² x + 1) dx = ∫(sec² x + cosec² x) dx = ∫(sec² x)dx + ∫(cosec² x)dx = tan x - cot x + C |
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| 18. |
\left. \begin{array} { c } { ( b + c - a ) \operatorname { tan } \frac { A } { 2 } = ( c + a - b ) \operatorname { tan } \frac { B } { 2 } } \\ { = ( a + b - c ) \operatorname { tan } \frac { C } { 2 } } \end{array} \right. |
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| 19. |
Let Us Practise 2.73. TandaFind the HCF of the following2 132 and 1651. 48 and 846. 96,64 and 15. 37 and 294. 1 and 14Find the greatest number that exactly divides 112 and 176.3 boys and 78 girls are to be divided into teams with equal number of memberan airls. What is the largest numbers |
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Answer» 1) 48 and 84; 48/2=24/2=12/2=6/2=3, 84/2=42/2=21/3=7;, 48=2×2×2×3; 84=2×2×3×7, 132 and 165; 132/2=66/2=33/11=3; 165x3=55/11=5; 2×2×3×11; 3×11; 1 and 7; 1 ,7 ;:; 1 and 14,; 14/2=7; 1, 2, 7, 14 37, 29; 1, 29, 37 HcF=1, 29, 37 96, 64,112; ; 96/2=48/2=24/2=12/2=6/2=3, 2x2x2x2x3; 64/2=32/2=16/2=8/2=4/2=2; 2x2x2x2x2x2; 112/2=56/2=28/2=14/2=7; 2x2x2x2x3; |
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| 20. |
sinx ka integration |
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Answer» Thanks |
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| 21. |
what is integration? |
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Answer» Ans :- Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. PLEASE LIKE THE ANSWER |
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| 22. |
and tan B =-, show that cos 2A = sin 4B.711. If tan A =3third |
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Answer» tanA = 1/7 and tanB = 1/3 show cos2A = sin4B sinA = 1/sqrt50 and cosA = 7/sqrt50 sinB = 1/sqrt10 and cosB = 3/sqrt10 now cos2A = 2cos^2A - 1 = 2{7/sqrt50}^2 -1 = 2* 49/50 = 49/25 -1=24/25 sin4B= 2 sin2B * cos2B = 4 sinB cosB {2 cos^2B - 1} = 4*1/sqrt10 3/sqrt10 [ 2 *9/10 - 1 ] =12/10 [4/5 ] = 24/25 so cos2A = sin4B proved |
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| 23. |
\frac { \operatorname { sin } ( A + B ) } { \operatorname { sin } ( A - B ) } = \frac { \alpha + \beta } { \alpha - \beta } , \text { show that } \frac { \operatorname { tan } A } { \operatorname { tan } B } = \frac { \alpha } { \beta } |
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| 24. |
A %2B B = 45 ^ \circ , \text Show that ( 1 %2B \operatorname tan A ) ( 1 %2B \operatorname tan B ) = 2 |
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Answer» Given A+B=45 {Take tan on both the sides } tan(A+B) = tan45 tanA+tanB/1- tanA tanB = 1 tanA+tanB=1-tanA.tanB tanA+tanB+tanA.tanB=1 adding "1" on both sides 1+ tanA+tanB+tanA.tanB=1+1 (1 + tanA)+tanB(1+tanA).=2 (1+tanA)(1+tanB)=2 Hence proved . |
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| 25. |
tana + tan Btan & tan ŕ¤ŕĽcota +cot |
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| 26. |
न रू ।सब्ध काजए केsin(A+B)= tanA+tan BcosA cosB |
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Answer» sin(a+B) / cosacosB = tana + tanB sin(a+B) = sinacosB + cosasinB (identity) sin(a+B) / cosacosB = (sinacosB + cosasinB) / (cosacosB) = sinacosB/cosacosB + cosasinB/cosacosB = sina/cosa + sinB/cosB = tana + tanB |
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| 27. |
tanattantIf A+B+C = T Prove thattan c = tana & tang & conc. |
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| 28. |
Prove that -| +tan?A) (l—tanA\2 . ,(l + cotzA) & (l ना पा दी-O] |
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Answer» L.H.S1+tan^2A/1+cot^2A=(1+sin^2A/cos2A)/(1+cos^2A/sin^2A)=((cos^2A+sin^2A)/cos^2A)/(sin^2A+cos^2A)/sin^2A=(1/cos^2A)/(1/sin^2A)=1/cos^2A*sin^2A/1=sin^2A/cos^2A=tan^2A M.H.S(1-tanA/1-cotA)^2=(1+tan^2A-2tanA)/(1+cot^2-2cotA)=(sec^2A-2tanA)/(cosec^2A-2cosA/sinA)=(sec^2A-2*sinA/cos A)/(cosec^2A-2cosA/sinA)=(1/cos^2A-2sinA/cosA)/(1/sin^2A-2cosA/sinA)=[(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin^2A]=(1-2sinAcosA)/cos^2A*sin^2A/(1-2sinAcosA)=sin^2A/cos^2A=tan^2A R.H.S=tan^2A |
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| 29. |
- tan A cotA20. Prove that —— .+ =~e 1-cotA 1-tanA =1+ SecAcosec A |
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Answer» LHS:tan A/(1 - cot A) + cot A /(1 - tan A) = (tan A)/[(1 - (1/tan A)] + cot A /(1 - tan A) = (tan^2 A)/[(tan A - 1)] + cot A /(1 - tan A) = (tan^2 A)/[(tan A - 1)] - cot A /(tan A - 1) = (tan^2 A - cot A) / (tan A - 1) = (tan^2 A - 1/tan A) / (tan A - 1) = (tan^3 A - 1) / [tan A (tan A - 1)] = (tan A - 1)(tan^2 A + tan A + 1) / [tan A (tan A - 1)] = (tan^2 A + tan A + 1) / tan A = 1 + tan A + cot A = 1 + [(sin A/cosA) + (cos A/sin A)] = 1 + [(sin^2 A + cos^2 A) / sin A cos A] = 1 + [1 / (sin A cos A)] = 1 + (sec A x cos A)= RHS Hence proved |
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| 30. |
Prove it -(1+tan? A) = (1-tanda = tanaIt lot? A) 1-Cotal |
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| 31. |
24.cos4x = 1 — 8sin“ x cos” x |
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Answer» cos4x ascos2(2x) =1-2sin²(2x) =1-2(2sinx . cosx)² Use { sin2x = 2sinx. cosx} =1-2(4sin²x.cos²x)=1-8sin²x.cos²xLHS=RHS |
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| 32. |
\operatorname { tan } A = \frac { 1 } { 7 } \text { and } \operatorname { tan } B = \frac { 1 } { 3 } , \text { show that cos } 2 A = \operatorname { sin } 4 B |
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| 33. |
👉If tan(A+B)=p,tan(A-B)=q;show that tan2A=p+q/1-pq.👈 |
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| 34. |
then the maximum value of /(x) is(a)e (b)(e((d) of these |
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| 35. |
cos4x=cos2x |
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Answer» Cos4x = cos2x Using formula : cos2a = 2cos²a - 1 2cos²2x - 1 = cos2x ⇒ 2cos²2x - cos2x -1 = 0⇒ 2cos²2x - 2cos2x + cos2x - 1 = 0⇒ 2cos2x(cos2x-1 ) + 1 (cos2x-1) = 0⇒ (cos2x-1)(2cos2x + 1) = 0⇒ cos2x = 1⇒ 2x = 2nπ +- 0 ⇒ x = nπ or cos2x = -1/2⇒ 2x = 2nπ +- 2π/3 ⇒ x = nπ +- π/3 Intersection : x ∈ nπ U nπ +- π/3 |
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| 36. |
integration of cos4x |
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| 37. |
1+cos4xcotx-tanx- dx |
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Answer» how it defined you r questions |
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| 38. |
8. Given a right-angled AABC; LC 90° in which3'tan B-_3 ; show that sin А .cos B-cos A.tanAsin B +2 |
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| 39. |
27.If x = 15°, then what is the value of 4 sin2r cos4x sinhx?ogFind the value |
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| 40. |
\frac{\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)}{\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right)}=\csc 2 \theta\frac{\sin (A+B)}{\sin (A-B)}=\frac{\alpha+\beta}{\alpha-\beta}, \text { show that } \frac{\tan A}{\tan B}=\frac{\alpha}{\beta} |
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Answer» i)use formula tan ((pi)/4 +x)=(1+tan x)/(1-tan x)and tan ((pi)/4-x)=(1-tan x)/(1+tan x) |
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| 41. |
24. Which term of the AP 121, 117, 113,is its first negative term? |
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Answer» Bhai pora question |
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| 42. |
tan ( A+B ) - tan B / 1+ tan ( A+ B ) tan B = tan A |
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| 43. |
4 Which term of the AP 121, 117, 113,is its first negative term?24. |
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| 44. |
16 If A and B are acute angles such that tan Atan B andtan (A +B) an A + tan Btan (AB)11- tan A tan B Show that A+B 45 |
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Answer» Using the formula for tan(A + B) given in the question |
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| 45. |
Differentiate:_x×√(1-cos2x)÷(1+cos2x) |
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Answer» Answer didn't match Answer given in the book is: xsec^2x+tanx |
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| 46. |
sin4x |
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| 47. |
15. For the AP a,аг, аз,if3, find12 37 |
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| 48. |
solve for x:cosx+cos2x+cos2x=0 |
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Answer» please somebody solve it sinx +cosx=1 |
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| 49. |
solve sin2x-sin4x+sin6x=0 |
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Answer» Sin2x-sin4x+sin6x=0 (sin6x+sn2x)-sin4x=0 We know that sinx +siny=2sin(x+Y/2)cos(x-y/2) Thereofre , =2sin(6x+2x/2)cos(6x-2x/2)-sin4x=0 =2sin8x/2cos4x/2-sin4x sin4x(2cos2x-1)=0 Therefore, sin4x=0 or 2cos2x-1=0 sin4x=0 or 2cos2x=1 sin4x=0 or cos2x=1/2
Now Let sinx=siny sin4x=sin4y now sin4x=0 Therfore sin4y=0 sin4y=sin(0) 4y=0 y=0 Now cos2x=1/2 Let cosx=cosy Cos2x=cos2y cos2y=1/2 ;cos(2y)=π/3;2y=π/3; Now, for sin4x=sin4y the general solution will be 4x=nπ±(-1)ⁿ4y Putting y=0 4x=nπ x=nπ/4 For cos2x=cos2y is 2x=2nπ±2y now 2y=π/3 2x=2nπ±π/3 on solving we get x=nπ±π/6 Therfore for sin4x=0 ,x=nπ/4 and for cos2x=1/2,x=nπ±π/6 |
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| 50. |
cos2x = 2sin^2y+4cos(x+y) sinxsiny +cos2x(x+y) |
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Answer» please mention what is to be done please mention what is to be done |
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