Saved Bookmarks
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. |
-sin(theta)^2 %2B tan(theta)^2=sin(theta)^2*tan(theta)^2 |
| Answer» | |
| 2. |
\sqrt { \frac { 1 + \operatorname { sin } \theta } { 1 - \operatorname { sin } \theta } } - \operatorname { sec } \theta = \operatorname { sec } \theta - \sqrt { \frac { 1 - \operatorname { sin } \theta } { 1 + \operatorname { sin } \theta } } |
| Answer» | |
| 3. |
\sin 2 x-\sin 4 x+\sin 6 x=0 |
| Answer» | |
| 4. |
\operatorname { sin } 2 x + 2 \operatorname { sin } 4 x + \operatorname { sin } 6 x = 4 \operatorname { cos } ^ { 2 } x \operatorname { sin } 4 x |
| Answer» | |
| 5. |
\frac { 1 } { \operatorname { cosec } \theta - \operatorname { cot } \theta } = \operatorname { cosec } \theta + \operatorname { cot } \theta |
|
Answer» in LHS multiply with cosec theta+cot thetaso it would be cosec theta+ cot theta/cosec^2theta-cot^2thetaas we know cosec^2 theta- cot^2 theta=1we got it same as RHs no |
|
| 6. |
sqrt((cosec*theta - 1)/(cosec*theta %2B 1))=(-sin(theta) %2B 1)/cos(theta) |
| Answer» | |
| 7. |
\operatorname { cos } ^ { 3 } 2 x + 3 \operatorname { cos } 2 x = 4 ( \operatorname { cos } ^ { 6 } x - \operatorname { sin } ^ { 6 } x ) |
| Answer» | |
| 8. |
\operatorname { sin } 2 x + \operatorname { sin } 4 x + \operatorname { sin } 6 x = 0 |
| Answer» | |
| 9. |
2 + 8 ( \operatorname { sin } ^ { 6 } x + \operatorname { cos } ^ { 6 } x ) = 26 |
| Answer» | |
| 10. |
( \sin \theta + \csc \theta ) ^ { 2 } + ( \cos \theta + \sec \theta ) ^ { 2 } = 7 + \tan ^ { 2 } \theta + \cot ^ { 2 } \theta |
| Answer» | |
| 11. |
\sqrt { \frac { 1 - \operatorname { sin } \theta } { 1 + \operatorname { sin } \theta } } + \sqrt { \frac { 1 + \operatorname { sin } \theta } { 1 - \operatorname { sin } \theta } } = 2 \operatorname { sec } \theta |
| Answer» | |
| 12. |
\sec ^ { 2 } \theta - \frac { \sin ^ { 2 } \theta - 2 \sin ^ { 4 } \theta } { 2 \cos ^ { 4 } \theta - \cos ^ { 2 } \theta } = 1 |
| Answer» | |
| 13. |
\frac { 2 \sin \theta - \sin 2 \theta } { 2 \sin \theta + \sin 2 \theta } = \tan ^ { 2 } \frac { \theta } { 2 } |
| Answer» | |
| 14. |
\left. \begin{array} { l } { \sqrt { 7 } \times \sqrt { 7 } = } \\ { \sqrt { 7 } + \sqrt { 7 } = } \end{array} \right. |
|
Answer» √7 x √7 = 7√7 + √7 = 2√7 |
|
| 15. |
2^(x - 2) %2B 5^(-2*x %2B 13)=2^(x %2B 2) %2B 5^(-2*x %2B 11) |
|
Answer» 5^13-2x+ 2^x-2=2^x+2+ 5^11-2x; 5^13-2x+5^;11-2x=2^x+2-2^x+2; 5^13-11-2x+2x=2^x+2'x-2; 5^2=25 |
|
| 16. |
\sqrt { 7 } - \frac { 3 } { 5 } \sqrt { 7 } + 2 \sqrt { 7 } |
| Answer» | |
| 17. |
Show that (\left(\frac{\sqrt{7}+i \sqrt{3}}{\sqrt{7}-i \sqrt{3}}+\frac{\sqrt{7}-i \sqrt{3}}{\sqrt{7}+i \sqrt{3}}\right)is real |
|
Answer» yo. gas are you preparing for jee |
|
| 18. |
tan(theta) %2B cot(theta)=sec(theta*(cosec*theta)) |
| Answer» | |
| 19. |
( \cosec \theta + \cot \theta ) ^ { 2 } = \frac { \sec \theta + 1 } { \sec \theta - 1 } |
| Answer» | |
| 20. |
\cosec\theta \tan \theta = \sec \theta |
|
Answer» cosec θ tan θ1/sinθ × (sinθ)/cosθ1/cosθsec(θ)
|
|
| 21. |
\sin 2 x %2B 2 \sin 4 x %2B \sin 6 x = 4 \cos ^ 2 x \cdot \sin |
|
Answer» L.H.S. = sin 2 x + 2 sin 4 x + sin 6 x= [sin 2 x + sin 6 x ] + 2 sin 4x= 2 sin 4 x cos 2 x + 2 sin 4x= 2 sin 4 x cos 2 x + 2 sin 4x= 2 sin 4 x (cos 2 x + 1)= 2sin4x(2cos x -1+1)= 2 sin 4 x (2 cos x )= 4cos x sin 4 x= R.H.S. |
|
| 22. |
sqrt(-cosec*theta %2B sec(theta))/sqrt(cosec*theta %2B sec(theta))=1/(sqrt(7)) |
| Answer» | |
| 23. |
\frac { \operatorname { cot } \theta } { \operatorname { cosec } \theta + 1 } + \frac { \operatorname { cosec } \theta + 1 } { \operatorname { cot } \theta } = 2 \operatorname { sec } \theta |
|
Answer» cota/(csca + 1) + (csca + 1)/cota =(csca - 1)/(csca - 1) * cota/(csca + 1) + (csca + 1)/cota =(csca cota - cota)/(csc^2a - 1) + (csca + 1)/cota =(csca cota - cota)/cot^2a + (csca + 1)/cota =(csca - 1)/cota + (csca + 1)/cota =2csca/cota =(2/sina)/(cosa/sina) =2/cosa =2seca |
|
| 24. |
\frac 2 \sin \theta 1 %2B \cos \theta %2B \sin \theta = \frac 1 - \cos \theta %2B \sin \theta 1 %2B \sin \theta |
|
Answer» Like if you find it useful |
|
| 25. |
\frac { \operatorname { tan } \theta } { \operatorname { sec } \theta - 1 } + \frac { \operatorname { tan } \theta } { \operatorname { sec } \theta + 1 } = 2 \operatorname { cosec } \theta |
| Answer» | |
| 26. |
sin(theta)^2 %2B sin(pi/6)^2 %2B sin(pi/3)^2 %2B sin(pi/2)^2 |
|
Answer» 2 is the correct answer to this question. |
|
| 27. |
((-sin(theta) %2B 1)*(2*sin(theta) %2B 2))/(((-2*cos(theta) %2B 2)*(cos(theta) %2B 1))) |
| Answer» | |
| 28. |
( \operatorname { sin } \theta + \operatorname { cosec } \theta ) ^ { 2 } = \operatorname { sin } ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta + 2 |
| Answer» | |
| 29. |
13.Find the value of k for which a-3b is a factorof a4 -7a2b2 +kb4. Hence, for this value offactorize a -7a2b2 +kb completely. |
| Answer» | |
| 30. |
0, show that al, arform an AP where a. is defined asbelow:(ii) as # 9-5n,Đ°",(i) a, 3+4n |
| Answer» | |
| 31. |
Show this defined as aat a1, a2, a3,an form A.Pwhere a,9-5n. |
|
Answer» Given an = 9-5nto know the series,a1 = 9-5(1) = 4a2 = 9-5(2) = 9-10 = -1a3 = 9-5(3) = 9-15 = -6a4 = 9-5(4) = 9-20 = -11 ......So the series is 4, -1, -6, -11 ...common difference d = a2 - a1 = -1-4 = -5So, the common difference d = -5 (we can check it by doing a3-a2, a4-a3 ... )when we do with the a3-a2 and a4-a3 too, we get the common difference d=-5So the {an} i.e, the series is in A.PThen, by usingSn = n/2 [2a +(n-1)d]In above Calculation,We got that a = 4, d = -5, n=15 (as per problem)*he wasnt mentioned that in question, he mentioned that in commentsS15 = 15/2 [ 2(4)+(15-1)(-5)]S15 = 15/2 [ 8+14(-5)]S15 = 15/2 (-70+8)S15 = 15/2 (-62)S15 = 15(-31)S15 = -465therefore the sum of the first 15 numbers on the series that we find out is -465 Thank You |
|
| 32. |
( 1 + \operatorname { cot } \theta + \operatorname { tan } \theta ) ( \operatorname { sin } \theta - \operatorname { cos } \theta ) = \frac { \operatorname { sec } \theta } { \operatorname { cosec } ^ { 2 } \theta } - \frac { \operatorname { cosec } \theta } { \operatorname { sec } ^ { 2 } \theta } |
| Answer» | |
| 33. |
Show that a, a,i) a, 3+4na, . form an AP where a, is defined as below:(i) a 9-5n |
|
Answer» 1)an=3+4n a1=3+4(1)=3+4=7 a2=3+4(2)=3+8=11 a3=3+4(3)=3+12=15 common difference=11-7=15-11=4first term=7A.P.=7,11,15....... |
|
| 34. |
\operatorname tan \theta = 1 \sqrt 7 , \text then \frac \operatorname cosec ^ 2 \theta - \operatorname sec ^ 2 \theta \operatorname cosec ^ 2 \theta %2B \operatorname sec ^ 2 \theta |
|
Answer» if tan theta = 1 / root 7cosec theta = root 8/1sec theta = root 8/root 7...........(cosec ^2 theta - sec ^2 theta )/cosec^2 theta +sec^2 theta )(root 8/1)^2 - (root 8/root 7)^2/ (root 8/1)^2 + (root 8/root 7)^2(8/1 - 8/7) / (8/1 + 8/7)((56 -8)/7 / ((56 + 8)/7[48/7 / 64/7][ 48 x 7]/ [7 x 64)Cancel 7's top and bottom=48/64Divide top and bottom by 16= 3/4 |
|
| 35. |
\operatorname { cosec } \theta = \frac { 3 } { 2 } , \text { then } 2 ( \operatorname { cosec } ^ { 2 } \theta + \operatorname { cot } ^ { 2 } \theta ) |
|
Answer» cosecx=3/2now cot^2x-cosec^2x=1so cotx=root(1+9/4)=root(13)/2so cosec^2x+cot^2x=(3/2)^2+(root(13)/2)^2 =9/4+13/4=22/4so 2(cosec^2x+cot^2x)=2*22/4=11 |
|
| 36. |
की 509 के ws + [yas a1 psos(e 2101g) hiepie BB 0Tpoos T |
|
Answer» LHS =cosA/(1+sinA) +(1+sinA)/cosA ={cos²A +(1+sinA)²}/cosA.(1+sinA) ={cos²A+1+sin²A+2sinA}/cosA.(1+sinA) use sin²∅ +cos²∅ =1 =(1+1+2sinA)/cosA(1+sinA) =2(1+sinA)/cosA(1+sinA) =2/cosA 2secA |
|
| 37. |
seca-1/seca+1=sin2a/(1+cosa)2=cosec theta sec theta-1/cosec theta +sec theta |
|
Answer» (secA – 1)/ (sec A + 1) (1/cosA - 1)/ (1/cos A +1) (1 – cosA/cosA)/ (1 + cosA/cosA) (1 – cosA)/ (1 + cosA) Multiplying numerator and denominator by (1 + cosA), we get (1 - cosA)(1 + cosA)/ (1 + cosA)2 (1 – cos^2A)/ (1 + cosA)^2 Sin^2A/ (1 + cosA)^2 [sinA/(1 + cosA)]^2= R.H.S Hence proved |
|
| 38. |
At midnight the water at a particular beach is at high tide. At the same time a gauge at thend of a pier reads 10 feet. Low tide is reached at 6 AM when the gauge reads 4ft.Determine the period, in hours. |
|
Answer» time period = 12-6= 6 hours |
|
| 39. |
0. Show that al, a" . . ., a", . . . form an AP where a is defined as below :(i) a = 3+4n(m) a, 9-5n |
|
Answer» i) a(n) = 3+4na(1) = 7a(2) = 11a(3) = 15so common difference is 4 , an AP therefore. ii) a(n) = 9-5na(1) = 4a(2) = -1a(3) = 9-15 = -6So common difference is -5, an AP therefore |
|
| 40. |
At midnight the water at a particular beach is at high tide. At the same time a gauge at theend of a pier reads 10 feet. Low tide is reached at 6 AM when the gauge reads 4ft.Determine the midline and write the answer as an equation.071405O 2 |
|
Answer» midline = 10+4/2= 7 |
|
| 41. |
प्रारूप4.1) 212+513 और 2-313 को जोड़ें। |
|
Answer» 3√2+2√3 is right answer |
|
| 42. |
Prove that (1+\cos \theta+i \sin \theta)^{n}+(1+\cos \theta-i \sin \theta)^{n}=2^{n+1} \cos ^{n}\left(\frac{\theta}{2}\right) \cos \left(\frac{n \theta}{2}\right) |
|
Answer» Using (cos(x)+i.sin(x))^n=(cos(nx)+i.sin(nx)) , we got the last step thank you for quick help |
|
| 43. |
(47) sin 75º = ?oh(13-1).2(d) 212d (12-1) |
|
Answer» sin75= sin(30+45)= (1/2 + 1/V2)=(V2+1)/2V2 sin(75) = sin ( 30 + 45) = sin(30 ) + sin ( 45); = 1/2 + 1 / V2 = = ( V2 + 1 / 2V2 ) |
|
| 44. |
cos(theta)/(cosec*theta %2B 1) %2B cos(theta)/(cosec*theta - 1)=2*tan(theta) |
|
Answer» cosx/ cosecx+1 + cosx/ cosecx-1=cosx( cosecx-1)/cosecx^2-1 + cosx( cosecx-1)/ cosecx^2+1= cosx(1/ sinx-1 sinx)+cosx(1/ sinx+1)=Tanx + tanx=2 tanx |
|
| 45. |
2. Evaluate the following:(a) 212-202 |
| Answer» | |
| 46. |
- 2Prove that | \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta} )^{2}=\frac{1-\cos \theta}{1+\cos \theta} |
|
Answer» LHS = [(1+sinA-cosA )/(1+sinA+cosA)]² =[ (1+sin²A+cos²A+2sinA-2sinAcosA-2cosA)/(1+sin²A+cos²A+2sinA+2sinAcosA+2cosA) ]² = [( 1+1+2sinA-2sinAcosA-2cosA)/(1+1+2sinA+2sinAcosA+2cosA)]² = { ( 2(1+sinA) -2cosA(sinA+1) / [ 2(1+sinA) + 2cosA(sinA +1 ) ]}² = { (2(1+sinA)[1 - cosA] /[2(1+sinA )(1 + cosA )] }² = [ ( 1 - cosA ) / ( 1 + cosA ) ]² = RHS |
|
| 47. |
1 %2B cot(theta)^2/(cosec*theta %2B 1)=cosec*theta |
| Answer» | |
| 48. |
\operatorname { cos } 2 \theta \operatorname { cos } \frac { \theta } { 2 } - \operatorname { cos } 3 \theta \operatorname { cos } \frac { 9 \theta } { 2 } = \operatorname { sin } 5 \theta \operatorname { sin } \frac { 5 \theta } { 2 } |
|
Answer» multiply and divide L.H.S. by 2. u'll get 1/2[2cos2x.cosx/2 - 2cos3x.cos9x/2] Now , 2cosAcosB = cos(A+B) + cos(A-B). so, LHS= 1/2[cos5x/2 + cos3x/2 - [cos15x/2 + cos3x/2] =1/2[cos5x/2 - cos15x/2] by using cosC - cosD= 2.sin(C+D)/2.sin(D-C)/2 therefore, LHS= 1/2[2.sin20x/4.sin10x/4] =sin5x.sin5x/2= RHS |
|
| 49. |
In âłABc P&.are the Midpoint ABand Bemid Point AB pier{ that auaare ĺąątue |
|
Answer» Let ABC is a triangle. P andQ are the mid points of AB and BC respectivelyand R is the mid-point of AP. Join PQ, QR, AQ, PC, RC as shown in thefigure. Now we know that median of a triangle divides it into two triangle of equal areas. In triangle CAP, Cr is the mid point. So Area(ΔCRA) = (1/2)* Area(ΔCAP) .......1 Again in triangle CAB, CP is the mid point. So Area(ΔCAP) = (1/2)* Area(ΔCPB) ............2 from eqaution 1 and 2, we get Area(ΔCAP) = (1/2)*Area(ΔCPB) ..............3 Again in triangle PBC, PQ is the mid point. So (1/2)*Area(ΔCPB) = Area(ΔPBQ) ............4 From equation 3 and 4, we get Area(ΔARC) = Area(ΔPBQ) ...............
|
|
| 50. |
tan(theta)/(sec(theta) %2B 1) %2B tan(theta)/(sec(theta) - 1)=2*(cosec*theta) |
|
Answer» sinx/cosx//(1-cosx/cosx) + sin x/cosx/(1+cosx/cosx)=sinx/(1-cosx)+sinx/(1+cosx)=sinx(1+cosx)+sinx(1-cosx)/(sinx)^2=sinx+sinx/(sinx)^2=2sinx /(sinx)^2=2/sin x=2cosecc |
|