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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. |
ACAB2 + BC2ABsee a-tan A = 1 + 2 tan 2 A.cos A cot A(2016)sec2A tan2 Aéż:LHS Scos A cot A4 se2A - tan A tan A |
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| 2. |
5. A shopkeeper has a stock of 140 kg of wheat. In the first four weeks, he sells 20 kg25 kg 26 kg and 47 kg of wheat respectively. How much wheat and what portion ofthe stock does he sell? |
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Answer» (20+25+26+47) + (1/2+3/4+1/2+1/4)= 118+2 =120 kg of wheat sell (140-120)/140 = 1/7 portion of stock does he sell |
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| 3. |
PapeWuite solution of the lintare lqunbin 5x toyinte vasca di |
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| 4. |
- SecB न सवारी |
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Answer» SecA+tanA=p ----------------------------(1)∵, sec²A-tan²A=1or, (secA+tanA)(secA-tanA)=1or, secA-tanA=1/p -----------------------(2)Subtracting (2) from (1) we get,2tanA=p-1/por, tanA=(p²-1)/2p∴, cotA=2p/(p²-1)Now, cosec²A-cot²A=1or, cosec²A=1+cot²Aor, cosec²A=1+{2p/(p²-1)}²or, cosec²A=1+4p²/(p²-1)²or, cosec²A=(p⁴-2p²+1+4p²)/(p²-1)²or, cosec²A=(p⁴+2p²+1)/(p²-1)²or, cosec²A=(p²+1)²/(p²-1)²or, cosecA=(p²+1)/(p²-1) |
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| 5. |
Prove that :tan θ + Cote -l-SecB cosecb |
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Answer» tanθ/(1 - cotθ) + cotθ/(1 - tanθ) => tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ) => tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1) => 1/(tanθ - 1) { tan²θ - 1/tanθ } => 1/(tanθ - 1) { (tan³θ - 1)/tanθ) [as, a³ - b³ = (a - b)(a² + b² + ab) => {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)} => tanθ + cotθ + 1 => sinθ/cosθ + cosθ/sinθ + 1 => (sin²θ + cos²θ)/sinθ . cosθ + 1 => 1/sinθ . cosθ + 1 => cosecθ . secθ + 1 please like the solution 👍 ✔️👍✔️ |
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| 6. |
If secB + tanθp, then find the value of cosec0. |
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Answer» Like if you find it useful |
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| 7. |
If secθ + tan8-p, then find the value of secB-tan? |
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| 8. |
30. | If SecB + tanθp, then find the value of cosecQ |
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Answer» Like if you find it useful |
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| 9. |
\begin{array}{l}{A+B=90^{\circ}, \text { prove that }} \\ {\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}=\tan A}\end{array} |
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Answer» Tan 1 . tan 2 .tan 3 . .......tan 88 . tan 89=sin1/cos1 .sin 2/cos 2 . ...sin 45/cos45..sin 88/cos 88 . sin 89/ cos 89=sin(90-89)/cos1 . sin(90-88)/cos2...........sin(90-45)/cos 45.........sin(90-1)/cos89=cos 89/cos1 .cos 88/cos 2......cos 45/cos 45 ..........cos 1/ cos 89 all the values will get cancelled......and the result will be 1 |
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| 10. |
Find out capital from the following informati(a) Total Assets 1,50,000(b) Creditors for goods90,000(c) Creditors for expenses 15,000(Ans.Capital is 45,000) |
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Answer» capital = total - (creditors for goods + creditors for expenses) capital = 150000 - (90000+15000)capital = 150000 - 105000= 45000 |
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| 11. |
Calculate total assets if Capital is Capital is 40,000, Creditors 230,000, Revenue earned durinthe period 275,000, Expenses incurred during the period 20,000.Value of Stock unsold20,000 |
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Answer» Capital after adjustments = Capital + Revenue - Expenses= 40000+75000+20000=1,35,000total assets= capital after adjustments+ creditors = 1,35,000+30000=165000 |
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| 12. |
Prove that (\tan A-\tan B)^{2}+(1+\tan A \tan B)^{2}=\sec ^{2} A \sec ^{2} B |
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| 13. |
\operatorname { tan } ^ { 2 } A - \operatorname { tan } ^ { 2 } B = \frac { \operatorname { sin } ^ { 2 } A - \operatorname { sin } ^ { 2 } B } { \operatorname { cos } ^ { 2 } A \operatorname { cos } ^ { 2 } B } |
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Answer» tan²x = sin²x/cos²x, and sin²x + cos²x = 1 tan²A - tan²B = sin²A/cos²A - sin²B/cos²B= sin²A cos²B/cos²A cos²B - sin²B cos²A/cos²A cos²B= sin²A(1-sin²B)/cos²A cos²B - sin²B(1-sin²A)/cos²A cos²B= (sin²A - sin²Asin²B)/cos²A cos²B - (sin²B - sin²Asin²B)/cos²A cos²B= (sin²A - sin²Asin²B) - (sin²B - sin²Asin²B)/cos²A cos²B= (sin²A - sin²Asin²B - sin²B + sin²Asin²B)/cos²A cos²B= (sin²A - sin²B)/cos²A cos²B |
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| 14. |
\begin{array} { l } { \text { 27. If } A + B = 90 ^ { \circ } \text { , prove that } } \\ { \sqrt { \frac { \tan A \tan B + \tan A \cot B } { \sin A \sec B } - \frac { \sin ^ { 2 } B } { \cos ^ { 2 } A } } = \tan A } \end{array} |
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| 15. |
\frac{\sec \theta}{\cos c c}+\frac{\cos e c \theta}{\sec \theta}=\sec \theta \cdot \cos \theta |
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| 16. |
1. Which of the following is correct(a) Assets(c) AssetsLiabilities + Capital ; (b) Assets - Capital - Liabilities;Liabilities - Capital ; (d) Assets External Equities |
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Answer» a) (a) |
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| 17. |
Complete the accounting equation(a) AssetsCapital1,00,000 = 80,000-Capital+ Liabilities+ Liabilities+ 40,000+ Creditors+(b) AssetsR2,00,000(c) Assets-Capital?!,60,00080,000 |
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Answer» The basic formula of accountancy is asset = capital + liability.a) liability = asset-capital = 20,000b) capital = asset - liability = 160000c) asset = 160000+80000 = 240000. |
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| 18. |
+Liabilities+Complete the accounting equation(a) AssetsCapital1,00,000 = 80,000(b) AssetsCapital2,00,000(c) AssetsCapital= 1,60,000+Liabilities+++*40,000Creditors80,000 |
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Answer» a),80,000+20,000; b)1,60,000, c)160000+80000=2,40,000 |
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| 19. |
1. Fill up the blanks on the basis of accounting equation:Assets Liabilities + Capital40,000 20,000 +- 10,000 + 15,00050,000.35,000 |
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Answer» 40000 = 20000 + 20000 25000 = 10000 + 15000 50000 = 15000 + 35000 |
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| 20. |
1)(C) nq(D) na, a,, аз an-l are in А.Р. with common difference d thennd [coseca, cosec a2 + cosecacosecat coseca3 coseca, t) ( y (C) sec a,-sec an terms]cot a,-cot a(B) tan a -tan a(D) cosen+1 |
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| 21. |
6 If tan28-1-a, prove that sec e+ tan'-(2-A2) 3/2cosec |
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| 22. |
22b tan, then prove that x2--#1.If xsec, ya2 b2 |
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Answer» Sec² A - tan² A = 1 Now , According to the problem given, x = a sec theta x / a = sec theta --------( 1 ) y = b tan theta y/b = tan theta -----------( 2 ) sec² theta - tan² theta = 1 ( x/a )² - ( y/b )² = 1 { from ( 1 ) & ( 2 ) } or b²x² - a² y² = a² b² |
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| 23. |
The centre of a circle is (2a, a - 7). Find the values of 'a' if the circle passes through thepoint (11, - 9) and has diameter 10/2 units. |
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| 24. |
The centre of a circle is (2a, a -7). Find the values of a if the circle passes through thepoint (11, -9) and has diameter 10 2 units |
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| 25. |
न संख्याओं का प्रणाम ज्ञात करने के लिए यूक्लिड विभाजन 0.15 और 225 (0 196 और 38220 (i) 867 शकT ST e o e o e |
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Answer» (1) prime factors of 135 and 225135 = 3*3*3*5225 = 3*3*5*5HCF = 3*3*5 = 45 (2) prime factors of 196 and 38220196 = 2*2*7*738220 = 2*2*3*5*7*7*13HCF = 2*2*7*7 = 196 (3) prime factors of 867 and 255867 = 3*17*17255 = 3*5*17HCF = 3*17 = 51 |
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| 26. |
1+sin® cos®29, सिद्ध करें : ८:56 * 7; ८6 * P |
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Answer» cos/(1 + sin) + (1 + sin)/cos = 2 sec common denominator is cos(1 + sin) [ cos ·cos + (1 + sin)(1 + sin) ] / [cos(1 + sin)] = 2 sec [ cos2+ sin2+ 2 sin + 1 ] / [cos(1 + sin)] = 2 sec [ 1 + 2 sin + 1] / [cos(1 + sin)] = 2 sec [ 2 sin + 2] /[cos(1 + sin)] = 2 sec [ 2(sin + 1] /[cos(1 + sin)] = 2 sec 2 1/cos = 2 sec 2 sec = 2sec Like my answer if you find it useful! |
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| 27. |
2) Which term of the following A. P. is 560?2, 1 1,20, 29,ä¸ |
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Answer» Let 560 is nth terma=2d=11-2=9 an=a+(n-1)d560=2+(n-1)9558/9=n-162=n-1n=62+1=63 |
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| 28. |
the rate oĹ R3 3.30 p29, If AABC is an isosccles triangle such that AB AC. Side BA is produced to D such thatAD AB. Show that angle BCD is a right angle. |
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| 29. |
m&: Ty- ÂĽ= STe0"S |
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| 30. |
A ittt st eFACTOR S2a%â qab - |
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Answer» 3a*a - 9ab = 3a (a - 3b) |
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| 31. |
12.1 B NIYHNIN O™ ३ ।॥ 0३ ०. ST e |
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| 32. |
Find the area of a rhombus whose baseand altitude are 1.2 m and 48 cm longrespectively |
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| 33. |
Thecentre of a circle is (2a, a-7). Find the values of a if the circle passes through thepoint (11, 9) and its diameter is 10 2 units. |
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| 34. |
If the circumference of a circle is Ď units more than thediameter of circle, then find the diameter of the circle. |
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Answer» Let diameter be d Given : pi d = pi + d So, pi d - d = pi d(pi - 1) = pi d = pi/(pi - 1) |
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| 35. |
2市何る戒倆tan θ = a'-b2af7 (a2-b2) sin θ + 2ab cos θ = a" + b",2ab |
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| 36. |
In the given figure AB and AC are opposite rays.(i) If x- 27, find y.(ii) If y 25°, find x(5y + 21)° |
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Answer» I) Sum of angles on a line = 180.=> 5y+21+2x = 180=> 5y+21+2(27)=180=> 5y + 21+54= 180.=> 5y = 180-75=> 5y = 105.=> y = 21. ii)=> 5y+21+2x = 180=> 5(25)+21+2x = 180=> 146+2x = 180=> 2x = 34=> X = 17. Please hit the like button if this helped you. thank you! |
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| 37. |
EXERCISE 7.31. In figure AB and AC are opposite rays1.. x=32°, find y2..y=28°, find x |
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Answer» from figure 2x+3y+11=180so 2x+3y=169 i) x=32 64+3y=169 so 3y=105 so y=35ii)y=28 2x+84=169 2x=85 so x=42.5 |
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| 38. |
16. In the given figure(10.19), OA and OB are opposite rays. Find thevalue of x. Further find ZAOC and LBOD.(x +10)(x +20)Fig. 10.19 |
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Answer» Sum of angles in straight line :x + 10 + x + x + 20 = 180°3x + 30 = 180°3x = 180° - 30°3x = 150°x = 150°/3x = 50°Angle AOC = 2x + 10 = 110°Angle BOD = x + 10 = 50° + 10 = 60° |
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| 39. |
के4oW e |
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Answer» x⁴ - ( y + z)⁴ (x²)² - (( y + z)²)² ( x² - (y+z)²) ( x² + (y+z)²) (x² - y² - z² - 2yz)× ( x² + y² + z² + 2yz) |
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| 40. |
sum of first three terms of a G. P. is 16 and the sum of next ths is 128. Determine, the first term, the common ratio and the sterms of the G. P.]29 |
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| 41. |
by G4. Ifthe first term of an AP is 22, the common difference is ㅢ and sum to-1n terms is 24, find the number of terms |
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Answer» a = 22 d = -4Sn = 24(n/2) (2a + (n - 1)d) = 24 n(44 - 4n + 4) = 48 48n - 4n^2 = 48 4n^2 - 48n + 48 = 0 n^2 - 12n + 12 = 0It cannot be factorised.Please recheck the question. please recheck the question. please recheck the question. |
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| 42. |
13. If the sum of the first 6 terms of a Gequal to 9 times the sum of its first 3determine its common ratio |
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| 43. |
2 cm wide, and 16 cm deep. If the box is open at the top.findr is 48 cm long, crunting it at the rate of 18rea of a cylinder is 5280 setop, find tu |
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| 44. |
>mmeet |
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Answer» 2m/3+8=m/2-12m/3-m/2=-1-8m/6=-9m=-9×6m=-54 Ans |
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| 45. |
| T Coer Y. = St e (etNS |
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Answer» tan²A + cot²A + 2 = tan²A + cot²A + 2 tan A × cot A = ( tan A + cot A)² = (sinA/cosA + cosA /sinA)² = ( sin²A + cos² A)²/ cos²A sin²A = sec²A cosec²A |
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| 46. |
15. A water tank is a right circular cylinder inshape. The diameter of its base 56 m and itsdepth is 10 m. Determine how many litres ofwater can it hold?(Take Ď=_ and 1 m3 = 1000 litres)mo- |
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Answer» Diameter of base of cylindrical tank = 56 m radius =562= 28 mDepth of cylindrical tank = 10 m volume of cylindrical tank=πr^2h=227×(28)^2×10=227×28×28×10=24640m^3 volume of water the tank can hold =24640×1000= 24640000 litres thank you so much Diameter =56 mradius=56/2m=28mdepth=10mvolume of cylinder =πrrh(22/7×28×28×10)m3= 24640m3Hence,1m3=1000litres24640×1000=24640000litres. |
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| 47. |
The centre of a circle is (2a, a-7). Find the value (s) of a, if the circle passes through the(11, -9) and has diameter 10 V2 units. |
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Answer» Diameter is 10√2then radius =5√2 √(2a-11)²+(a-7+9)²=5√2(2a-11)²+(a+2)²=(5√2)²4a²-44a+121+a²+4a+4=505a²-40a+75=0a²-8a+15=0(a-3)(a-5)=0a=3,5 |
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| 48. |
(d) If tan A=v2-1 then show thatSin A Cos A=v2/4 |
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| 49. |
1/3πr×rh |
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Answer» This is the formulae of volume of cone. |
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| 50. |
zCj_{ X rh) &hyjders |
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Answer» (4-¹×3-¹)²=(1/4 × 1/3)²=1/16 × 1/9=1/144 (4-1x3-1)2=12-2=144-1=1/144 (1/4×1/3)^2=(1/12)^2=1/144 |
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