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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. |
Find the value of k for which a -3bis a factorof a4 -7a2b2 +kb". Hence, for this value offactorize a4 -7a b2 +kb4 completely212 |
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Answer» thnx |
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| 2. |
ही ही. the, elmade \md Qe with ¥— axis |
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Answer» slope = tan = √3 We know, tan 30° = √3 = 30° angle makes with x axis |
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| 3. |
md the amount for75,000 at 896 fmannumCompounded annuafox too yeas s |
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Answer» A = 15000(1+8/100)(1+8/100) = 15000(1.08)(1.08) = 17496 |
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| 4. |
1—tan’45 नि नर नX 1+ tan*A4|= c0s?A4 — sin’4 |
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| 5. |
\begin { equation } \begin{array}{l}{\text { If } \sin \theta+\sin ^{2} \theta+\sin ^{3} \theta=1, \text { then prove that }} \\ {\cos ^{6} \theta-4 \cos ^{4} \theta+8 \cos ^{2} \theta=4}\end{array} \end { equation } |
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| 6. |
\left. \begin{array} { l } { ( i ) \operatorname { sin } ^ { 4 } \theta - \operatorname { cos } ^ { 4 } \theta = \operatorname { sin } ^ { 2 } \theta - \operatorname { cos } ^ { 2 } \theta } \\ { \operatorname { tan } ^ { 4 } \theta + \operatorname { tan } ^ { 2 } \theta = \operatorname { sec } ^ { 4 } \theta - \operatorname { sec } ^ { 2 } \theta } \end{array} \right. |
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Answer» 1) sin²∅+cos²∅ = 1 so, sin⁴∅-cos⁴∅ = (sin²∅+cos²∅)(sin²∅-cos²∅) = 1*(sin²∅-cos²∅) 2) sec²∅-tan²∅ = 1 tan⁴∅+tan²∅ =sec⁴∅-sec²∅=> sec⁴∅-tan⁴∅= tan²∅+sec²∅=> (sec²∅-tan²∅)(sec²∅+tan²∅)= tan²∅+sec²∅=> 1*(sec²∅+tan²∅) = tan²∅+sec²∅ |
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| 7. |
If cosec θ-cot θ cot θ, then prove thatcosec θ + cot θ--/2 cosec θ. |
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| 8. |
(sin(theta)^4 - cos(theta)^4)/(sin(theta)^2 - cos(theta)^2)=1 |
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| 9. |
cosec-g-1cosec θ |
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Answer» cosec^2-1=cot^2aHence √cot^2a/coseca=cota/coseca=cosa/sina*sina=cosAhence proved. |
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| 10. |
sin A+cos A4 S sin4-cosd _ 2L =S डयve tha sin A+cosAd sin’ A-cos 4DMLsin A —cos 4 |
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Answer» (sinA+cosA)^2+(sinA-cosA)^2/ sin^2A-cos^2A sin^2A+cos^2A+sin^2A+cos^2A / sin^2A-cos^2A 2/sin^2A-cos^2A |
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| 11. |
Hr (cose A + sin A) (cosec A -sin A)cosec A sin A) (cosec- cot2 A + cos A |
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| 12. |
(h)(a4-b") รท (a2 + b2) |
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| 13. |
Iand cos A 4)-O.A B,md A4,a |
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| 14. |
(cosec²A-1)cosec A |
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| 15. |
EXAMPLE 1.1.If a = x.j!, b = x.yt-, c = x.you? then show that a4 b? 24 = 1. |
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Answer» very tough question. you should ask you teacher. |
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| 16. |
2A4 Match the foilowings pairs of symomyms2) Gather3) Introduced4) impactA) ColectB) OrgunisedC) ImpressionD) ConductedE) Expreasion |
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Answer» 1. B2. A3. D4. Care the answers |
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| 17. |
sin(theta)^4 %2B 2*(sin(theta)^2*cos(theta)^2) %2B cos(theta)^4=1 |
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Answer» Let theta = x LHS : sin^4 x + 2sin^2 x cos^2 x + cos^4 x = (sin^2 x)^2 + 2sin^2 x cos^2 x + (cos^2 x)^2 = (sin^2 x + cos^2 x)^2 [Using sin^2 x + cos^2 x = 1] = 1 = RHS Hence proved To prove:sin⁴ theta+2sin²theta cos ²theta+cos ⁴theta =1 LHS:sin⁴ theta+2sin²theta cos ²theta+cos ⁴theta =(sin²theta+cos ²theta)² =(1)² =1 |
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| 18. |
(cosec^2*theta)*(sin(theta)^4 - cos(theta)^4 %2B 1)=2 |
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Answer» (sin⁴θ-cos⁴θ+1)cosec²θ=[{(sin²θ)²-(cos²θ)²}+1]cosec²θ=[{(sin²θ+cos²θ)(sin²θ-cos²θ)}+1]cosec²θ=(sin²θ-cos²θ+1)cosec²θ [Using sin²θ+cos²θ=1]={sin²θ+(1-cos²θ)}cosec²θ=(sin²θ+sin²θ)cosec²θ=2sin²θ.cosec²θ=2sin²θ×1/sin²θ=2 (Proved) (sin4θ - cos4θ + 1) cosec²θ =[ (sin²θ+cos²θ) (sin²θ-cos²θ) + 1] cosec²θ =[1 (sin²θ - cos²θ) +1] cosec²θ =[ sin²θ - (1-sin²θ) +1] cosec²θ =(sin²θ -1 +sin²θ +1 )cosec²θ =2sin²θ cosec²θ=2 =R.H.S HENCE PROVED |
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| 19. |
4 \tan \theta=3, \text { evaluate }\left(\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}\right) |
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| 20. |
If $ 4 \tan \theta=3 $ Evaluate $ \left[\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}\right] $ |
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| 21. |
If sin theta=frac{4}{5},then find frac{4tan theta-5cos theta}{sec theta+4 cot theta} |
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| 22. |
lfsecθ + tanθ--p, then find the value of cosec6. |
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Answer» Like if you find it useful |
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| 23. |
If A + B+ C 180 ° then prove thattan A + tan B + tan C is equal to 1tan B tan C |
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Answer» Please like the solution 👍 ✔️ |
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| 24. |
In ΔΑ1C, prove that(i)tan A + tan B .i tan Ctan A tan B tan CA B C2cot--+ cot,i. cot |
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Answer» We have to prove that tan A + tan B + tan C = tan A*tan B*tan C for any non-right angle triangle.For any triangle the sum of the angles is equal to 180 degrees. If we take a triangle ABC, A + B + C = 180 degrees. A + B + C = 180 or A + B = 180 - Ctan (A + B) = tan (180 - C) = tan C=> (tan A + tan B)/(1 - tan A*tan B) = tan C=> tan A + tan B = tan C - tan A*tan B*tan C=> tan A + tan B + tan C = tan A*tan B*tan CThis proves that tan A + tan B + tan C = tan A*tan B*tan C next part |
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| 25. |
âł ABC01BO8 |
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| 26. |
\begin{array} { l } { \cdot \text { In a } \Delta A B C \text { , prove that } } \\ { \tan A + \tan B + \tan C = \tan A \tan B \tan C } \end{array} |
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Answer» Solution :- For any triangle the sum of the angles is equal to 180 degrees. If we take a triangle ABC, A + B + C = 180 degrees. A + B + C = 180 or A + B = 180 - C tan (A + B) = tan (180 - C) = tan C => (tan A + tan B)/(1 - tan A*tan B) = tan C => tan A + tan B = tan C - tan A*tan B*tan C => tan A + tan B + tan C = tan A*tan B*tan C |
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| 27. |
Cosec A /cosec A - 1 +cosec A /cosec A +1=2 sec²A |
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| 28. |
17. If A, B, C are angles of a triangle, prove thatA+B(ii) cos A4 B = sin(ii) cos(1) sin (A + B) = sin Ctan (B+C) + tan (C+ A) + tan (A+B)tan (It - A) + tan (2 - B) + tan (31-C) |
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| 29. |
Cosec A/cosec A -1 + cosec A/cosec A + 1 |
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| 30. |
\begin { equation } \frac{\cosec \theta}{(\cosec \theta-1)}+\frac{\cosec \theta}{(\cosec \theta+1)}=2 \sec ^{2} \theta \end { equation } |
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| 31. |
cosec 4 cosec 6â = sec?cosecâ1 cosecO+1 |
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| 32. |
If π/2 < θ < π then the distance between the points (cot θ, 3), (0,2) is1 ) sec θ2) cosec θ3)-sec θ4) -cosec θ |
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| 33. |
cot(theta)^4 - 1=cosec^4*theta - 2*cosec^2*theta |
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Answer» Let theta = A LHS = cot^4A-1 = (cot^2A)^2 - 1^2 = (cot^2A + 1)(cot^2A - 1) = cosec^2A(cosec^2A - 1 - 1) [since 1+cot^2A = cosec^2A which implies cot^2A = cosec^2A-1] = cosec^2A(cosec^2A - 2) = cosec^4A - 2cosec^2A = RHS Hence proved |
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| 34. |
cot A +cosec A-1 पलट न4. il oo t A—cosec A+1 |
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Answer» Cosec A + cot A - 1 / cot A - cosec A + 1we know that,cosec ² A - cot ² A = 1substituting this in the numeratorcosec A + cot A -(cosec ² A - cot ² A) / (cot A - cosec A + 1)x²-y²= (x+y)(x-y)cosec A + cot A - (cosec A + cot A) (cosec A - cot A) / (cot A - cosec A + 1)taking common(cosec A + cot A)(1-cosec A + cot A) / (cot A - cosec A + 1)cancelling like terms in numerator and denominatorwe are left with cosec A + cot A= 1/sin A + cos A/sin A= (1+cos A) / sin A |
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| 35. |
If π/2 < θ < π then the distance between the points (cot θ, 3), (0, 2) is1)sec θ2) cosec θ3)-sec θ4) cosec θ |
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| 36. |
tan A(1 - cot A) (1 -tan A)cot A=(1 + tan A + cot A).s |
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| 37. |
1. यदि sin 3A = cos (A - 26°), जहाँ 3A एक न्यून कोण है, A का मान ज्ञात कीजिए। |
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Answer» sin 3A=cos(A −26)⇒cos(90− 3A) = cos(A−26)⇒(90−3A) = (A−26)⇒4A = 116°∴ A = 29° |
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| 38. |
15, A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should itsspeed be increased so that it may take only 8 hours to cover the same distance? |
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Answer» 12km/hr |
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| 39. |
A train leaves station X at 5 a.m. and reaches station Y at 9a.m. Another train leaves station Y at 7 a.m. and reachesstation X at 10: 30 a.m. At what time do the two trains crosseach other ?(a) 7:36 am(c) 8:36 am(b) 7:56 am(d) 8:56 am |
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Answer» Time taken by train 1 from x to y = 4 hours It travels for 2 hours before train 2 starts. Time taken by train 2 from y to x = 3.5 hours Let distance between x & y = LCM(4,3.5) = 28 kms Speed of Train 1 = 28/4 = 7 kmph Speed of Train 2 = 28/3.5 = 8 kmph Train 1 covers 7 x 2 = 14 km by the time Train 2 starts at 7 AM. So, distance between the two trains at 7 AM is 28 - 14 = 14km Relative speed = 7 + 8 = 15 kmph Time taken to cross each other = 14/15 hours or 56 minutes. => They meet at 7:56 AM. (b) is correct option. |
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| 40. |
A train leaves point A at 7.00 a.m., travels180 kms and reaches point B at 10.00 a.m.What is the speed of the train? |
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| 41. |
ulativeRevisionress each of the following as ratio of the first quantity to the second, in its lowest term:in the form a:b, (ii) as a fraction25 paise, 50 paise2. 210°, 360°3. 250 cm, 1 m4. 80 paise, 1.206. 3 minutes 30 seconds, 1 houry and complete the following equivalent ratios:2:3 = 6:9 2. 3:8=12:32 3. 6:24= 3:12 4. 12: 7 = 36:211 kg 250 g, 3 kgd the value of the unknown in each:27Saco7 392. 69V? 12py and complete the following:ypes 900 words in 1 hour, his rate of typing isNICO4.7-86 ?If a typist types 900 WIf a man pays 6000 rent for 3 m1*300.00 is charged forA machine is used toharged for 75 units of electricity thertoiwords per minute.rupees per month |
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Answer» (i)15 words in one minute (ii)(6000/3)rupees=2000rupees 1)15 words per minute 1)15 words per minutes2)2000 per month 15 words in one minute bcoz 900/60=15 |
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| 42. |
17. A train covers 101 km in 1 hour.How much distance will it cover in 4 hours ?5 |
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| 43. |
EXERCISE17(A)1. A train covers 51 km in 3 hours. Calculate its speed. How far does the train go in30 minutes? |
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| 44. |
10. A can do of a certain work in 16 days and B can do- of the same work in 3 days. In howmany days can both finish the work, working together? |
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| 45. |
4 4 >: 2-0; 1 ने 1-2 0057 3 ने.का -~ (viii) sin? A - cos |
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| 46. |
4) No rght angles1os |
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Answer» since the two sides of triangle are equal.. so the lower base angles are equal the upper angle is 180-110 => x+x + (180-110) = 180=> 2x + 70 = 180=> 2x = 180-70 = 110=> x = 110/2 = 55° |
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| 47. |
(ii) sin C, cos C2 In Fig. 8.13, find tan P- cot R-cot R.34ré sin Acalculate cos A and tan A.12 em13 em4 Given 15 cot A -8, find sin A and sec A.13calculate all other trigonometric ratios.Fig. 8.13 |
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Answer» in fig 8.13 ,find tan p_cotR. |
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| 48. |
(f)) (2ry +5y |
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Answer» (2xy+5y)^2using the identity (a+b)^2=a^2+b^2+2abhence4x^2y^2+25y^2+20xy^2 |
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| 49. |
Prove that\begin{aligned}\left(\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}\right)\left(\frac{\sin A}{1-\cos A}-\frac{1-\cos A}{\sin A}\right) \\=& 4 \cosec A \cot A \end{aligned} % 1+ c__osA sin AA sin A 1 - co;=4 cosec A cot . |
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Answer» Please send it once again Thanks |
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| 50. |
Evaluate , 3 sin 3A+ 2 coS (2A5) when A 202 cos 3A -sin (2A10°) |
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