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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. |
9.Findthe period of sin4x + cos4xSolution. |
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| 2. |
findhevalseoC.e value band |
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Answer» put in p(1)(1)^2-3(1)+11-3+1-2+1-1and for p(-1)(-1)^2-3(-1)+11+3+15 |
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| 3. |
17. Show that the relationin the strRH(a,b): both a and bare either even or odd) ,show that R is an equivalence relation. furthershow that all the elements of set (1,3,5,7) are related to each other and all the elements of set{2,4,6} are related to each other but no elements of set {1,3,5,7} is related to any elements of set |
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Answer» ABCD is trapezium in which AB||CD, show that A=B and C=D |
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| 4. |
tan A+tan BShow thattan A +cotB= tanA . tan BB |
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Answer» There must be an error in the question so this would be the answer (tan A+tan B) / (cot A+cot B)⇒(tan A+tan B) / (1/tan A+1/tan B)⇒(tan A +tan B) / (tan A+tan B / tan A·tan B)⇒(tan A+tan B) / (tan A·tan B / tan A+tan B)⇒tan A·tan Bhence proved |
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| 5. |
EnglichQuestion No. 1Select the number that is related to the Hird mumber in the sume way as the second sumber is related to the first mmb121:21 100:?101219 |
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| 6. |
\operatorname { sin } ( A - B ) ( \operatorname { tan } A + \operatorname { tan } B ) = \operatorname { sin } ( A + B ) ( \operatorname { tan } A - \operatorname { tan } B |
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Answer» Sin(A-B) (TanA+TanB) =Sin(A+B) (TanA-TanB) Taking L. H. S. first=(SinA-SinB) (SinA/CosA+SinB/CosB) =Sin^2A/CosA-Sin^2B/CosBNow taking R. H. S.... =(SinA+SinB) (SinA/CosA-SinB/CosB) =Sin^2A/CosA-Sin^2B/CosBHence.. L. H. S=R. H. S. proved r u atupid stupid |
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| 7. |
\frac { \tan A + \tan B } { \cot A + \cot B } = \tan A \tan B |
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Answer» tanA + tanB------------------- cotA + cotB tanA + tanB---------------------------1/tanA + 1/tanB tanA + tanB------------------ × tanAtanB = tanAtanBtanA+ tanB thanks bhai |
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| 8. |
3. Find the sum of each of the following.(i) 7a +6b, -3a + 9b and 6a-3b |
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Answer» 7a+6b-3a+9b+6a-3b = 10a+12b |
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| 9. |
\cot \left( \frac { \pi } { 4 } + x \right) \cdot \cot \left( \frac { \pi } { 4 } - x \right) = 1 |
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| 10. |
\frac { \operatorname { cot } A + \operatorname { tan } B } { \operatorname { cot } B + \operatorname { tan } A } = \operatorname { cot } A \operatorname { tan } B |
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Answer» cotA=1/tanAcotB=1/tanBputting in the eq.=(1/tanA +tanB)/(1/tanB +tanA)=((1+tanAtanB)/tanA)/((1+tanAtanB)/tanB)=tanB/tanA=cotA tanB |
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| 11. |
\frac { \operatorname { tan } A } { ( 1 - \operatorname { cot } A ) } + \frac { \operatorname { cot } A } { ( 1 - \operatorname { tan } A ) } = ( 1 + \operatorname { tan } A + \operatorname { cot } A ) |
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| 12. |
Show that (1+tan'd)-(1+tant4)2 = tanA= tan"Ahow that (1+Cot27)= (1+cot 4) |
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| 13. |
\csc ^{4} \theta-\csc ^{2} \theta=\cot ^{4} \theta+\cot ^{2} |
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Answer» cosec⁴θ - cosec²θcosec²θ ( cosec²θ - 1)(1 + cot²θ) cot²θcot²θ + cot⁴θProved |
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| 14. |
A + B = \frac { \pi } { 4 } , \text { show that (cot } A - 1 ) ( \operatorname { cot } B - 1 ) = 2 |
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Answer» Thanks bhai |
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| 15. |
\cot ^ { 2 } 30 ^ { \circ } - 2 \cos ^ { 2 } 60 ^ { \circ } - \frac { 3 } { 4 } \sin ^ { 2 } 45 ^ { \circ } - 4 \sin ^ { 2 } 30 ^ { \circ } |
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Answer» (root(3))^2-2(1/4)-3/4(1/root(2))^2-4(1/2)^2=3-1/2-3/8-1=21/8-3/2=(21-12)/8=9/8 |
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| 16. |
Prove that \frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x |
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| 17. |
For the AP:-3,-7, -11, ., can we find directly ay- a without actually findinga and a? Give reasons for your answer.302020 |
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Answer» here a= -3d= -7+3 = -4a(n) = a + (n-1)da(30) = a + 29×(-4)a(20) = a + 19 ×(-4)a(30) - a(20) = a + 29×(-4) - a + 19×4= -116 + 76 = -40 Here d = -4 hence your answer is wrong Thanks for letting us know |
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| 18. |
\left. \begin{array} { l } { 3 ^ { \circ } = p , \text { write } \operatorname { cos } 43 ^ { \circ } \text { in terms } } \\ { 0 ^ { \circ } = a , \text { write } \operatorname { sin } 50 ^ { \circ } \text { in terms } } \\ { \text { e value of } \operatorname { cos } ^ { 2 } 72 ^ { \circ } + \operatorname { sin } ^ { 2 } 72 ^ { \circ } } \end{array} \right. |
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Answer» a)sin43= pcos43= √1-p^2b)cot50=a=costheta/sinthetasin50= 1/√1+a^2 c) 1 |
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| 19. |
U2.You know that = = 0.142857. Can you predict what the decimal expansions of -47'57'67are, without actually doing the long division? If so, how?caHint : Study the remainders while finding the value of |
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Answer» 2/7=0.285714,3/7=0.428571,4/7=0.571428,5/7=0.714285,6/7=0.857142 |
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| 20. |
2.Write the acute angle θ satisfyingV3 sin θ=cos θ |
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| 21. |
\csc ^{4} \theta-\csc ^{2} \theta=\cot ^{4} \theta+\cot ^{2} \theta |
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| 22. |
2.ABC has a right angle at A. AB = 20, AC = 21, BC = 29.Write down sin B, cos C and tan B. |
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| 23. |
Obtain the volume of rectangular boxes with the following length, breadth and heirespectivelyo 5a, 3a', 7a(i) 2p, 4q, 8r (ii) xy, 2r'y, 2xy(v) a. 2b. 3 |
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| 24. |
\operatorname { cosec } ^ { 4 } x - \operatorname { cosec } ^ { 2 } x = \operatorname { cot } ^ { 4 } x + \operatorname { cot } ^ { 2 } x |
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Answer» cosec^4 x- cosec^2 x = ( cot^2 x + 1 )^2 - (cot^2 x +1)= cot^4 A + cot^2 A |
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| 25. |
18. In the figure, AABC is an obtuse triangle in which ZC >90°. AD L BOproduced and BEä¸AC produced. Prove thatAB2 = BC-BD + AC-AE. |
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| 26. |
\operatorname cosec ^ 4 A - \operatorname cosec ^ 2 A = \operatorname cot ^ 4 A %2B \operatorname cot ^ 2 A |
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Answer» To prove cosec⁴ A - cosec²A=cot⁴ A+cot²Asolving LHSLHS=cosec⁴x -cosec²x=cosec²x(cosec²-1)=cosec²xcot²x=(cot²x+1)cot²x=cot⁴x+cot²x proved |
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| 27. |
In Fig. 6.40, E is a point on side Cproduced of an isosceles triangle ABCwith AB-AC. If AD L BC and EF LAC,prove that Δ ABD-AECF. |
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| 28. |
(iv) cot-1 cot4 |
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Answer» (5pi/4) = pi + pi/4cot(pi + pi/4) = cot (pi/4)cot^-1[cot (pi/4) ] = pi/4 PLEASE HIT THE LIKE BUTTON |
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| 29. |
2 \sec ^{2} \theta-\sec ^{4} \theta-2 \csc ^{2} \theta+\csc ^{4} \theta=\cot ^{4} \theta-\tan ^{4} \theta |
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| 30. |
\left. \begin{array} { l } { \text { If } \quad \operatorname { tan } x + \operatorname { cot } x = 3 } \\ { \operatorname { tan } ^ { 4 } x + \operatorname { cot } ^ { 4 } x = 47 } \end{array} \right. |
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| 31. |
Aan-sin2(C) cot A sin(D) cot 4 sin A |
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Answer» option A is correct. |
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| 32. |
B+C4.In a triangleABC, write cos|ä¸2ě in terms of angle A. |
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Answer» A+B+C = 180 (ASP of trianlge) ~ B + C = 180 - A ~ cos (B + C) / 2 = cos (180 - A) / 2 ~ cos (180/2 - A/2) ~ cos (90 - A/2) ~ sin (A/2) Therefore, cos (B + C/2) = sin (A/2) Like my answer if you find it useful! |
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| 33. |
\frac { 1 } { \operatorname { tan } 3 A + \operatorname { tan } A } - \frac { 1 } { \operatorname { cot } 3 A + \operatorname { cot } A } = \operatorname { cot } 4 A |
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| 34. |
B + CIn a triangle ABC, write cos |! , in terms of angle A. |
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Answer» A+B+C = 180 (ASP of trianlge) ~ B + C = 180 - A ~ cos (B + C) / 2 = cos (180 - A) / 2 ~ cos (180/2 - A/2)~ cos (90 - A/2) ~ sin (A/2) Therefore, cos (B + C/2) = sin (A/2) |
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| 35. |
C.B+ CIn a triangle ABC, write cosin terms of angle A. |
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Answer» In ∆ we have A + B + C = 180° => ( B + C)/2 = 90 - A/2 => cos ( B + C)/2 = cos ( 90 - A/2) = sin A/2 |
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| 36. |
Q. 24.Without finding the cubes, factorise (x-2y)3 + (2y-3z)3 + (3-x)3ORFind the product (2x-y+ 32) (4x2+y+922 +2ry + 3yz - 6xz). |
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| 37. |
In a triangle ABC, write \cos \left(\frac{B+C}{2}\right) in terms of angle A. |
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| 38. |
\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta} |
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| 39. |
rectangular boxes with the following length, breadth and heigh2p, 4g. 8 (iy, 2ry, 2xy (iv) a, 2b, 3c4.Obtain the volume ofrespectivelywith thea |
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| 40. |
Prove that8tanA+tan A+ cot A). (CBSE 2010)cot A1-сot A' (1-tan A) - (1+tan A + Cot |
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Answer» tanA/ 1- cotA + cotA/1- tanA = sina/ cosa / 1- cosa/ sinA +^ cosa/ sina/ 1 - sina/ cosa = sina/ cosa/ sina- cosa/ sinA + cosa/ sinA/ cosa - sina/ cosa =sina^2/ cosa( sina- cosa) + cosa^2/ sina( cosa - sina) = sina^2+ cosa^2/ sinacosa= sina^2/ cosa sina- sina^2 + cosa^2/ sina cosa- cosa^2 = 1 + sinacosa/ sina cosa = 1/ sina cosa + |
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| 41. |
Fig. 10.12Fig. 10.13In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre 0 andanother tangent AB with point of contact C intersecting XY at A and X'Y' at B. Provethat L AOB = 90°. |
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| 42. |
numbers of houses preceding the house number X is cqual to sum of thc houses following t.2s. In figure, M is the mid-point of side CD of a parailelopran ABCD. The linc BM is drawn intersectingL. and AD produced st L. Prove tlhat I-2131A. |
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Answer» Like my answer if you find it useful! |
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| 43. |
Prove that[1/(cosec-cot ) ]-[1/sin]=[1/sin]-[1/(cosec+cot)] |
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| 44. |
1 (1) 1+ cosec Acosec Acos2 A1-sin A |
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Answer» LHS= RHS is the correct answer of this question |
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| 45. |
(cosec A-1), (cosec A + I)21, (1 + cot 4-cosec A) (1 + tan A + sec A) = 2A)2/+ (cos A + sec A)2-7 + tan2 A +2 A |
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| 46. |
B+C(viii) In a AABC, write cosin terms of angle A. |
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Answer» A+B+C = 180 (ASP of trianlge) ~ B + C = 180 - A ~ cos (B + C) / 2 = cos (180 - A) / 2 ~ cos (180/2 - A/2)~ cos (90 - A/2) ~ sin (A/2) Therefore, cos (B + C/2) = sin (A/2) |
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| 47. |
(vi) $9 x^{2} y^{2}-16 \quad$ (vii) $\left(x^{2}-2 x y+y^{2}\right)-z^{2}$(viii) $25 a^{2}-4 b^{2}+28 b c-49 c^{2}$ |
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Answer» vi) 9x^2y^2-16=(3xy+4)(3xy-4)vii) (x-y)^2-z^2=(x-y+z)(x-y-z)viii) (5a)^2-(2b-7c)^2=(5a-2b+7c)(6a+2b-7c) |
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| 48. |
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Pro0 is equidistant from A, B and C.(Hint: Extend BO to D such that BO = OD. Join A to D and C to D) |
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Answer» thanks😊😊😊😊😊🌺🌺i need more answer |
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| 49. |
PR, S 8 PR T SN3 यदि A+B+ C =90° a‘ सिद्ध कीजिए-C +tanC tan A =1<3 B s e तकहरणtan A tan B + tan B tan |
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Answer» A + B + C = 90° → B + C = 90°−A and tan (B+C)=tan(90°−A) and (tanB+tanC)/(1−tanBtanC)= cotA=cosA/sinA (tanB+tanC)/(1−tanBtanC)= 1/tanAor (tanB+tanC)tanA= (1−tanBtanC) →tanAtanB+tanAtanC =1−tanBtanC hence tanAtanB+tanAtanC+ tanBtanC=1 or tanAtanB+tanBtanC+ tanCtanA=1 Like my answer if you find it useful! |
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| 50. |
0. | If secθ + tanθ = p, then find the value of cosec6. |
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Answer» Secθ+tanθ=p ----------------------(1)∵, sec²θ-tan²θ=1or, (secθ+tanθ)(secθ-tanθ)=1or, secθ-tanθ=1/p ----------------(2)Adding (1) and (2) we get,2secθ=p+1/por, secθ=(p²+1)/2p∴, cosθ=1/secθ=2p/(p²+1)∴, sinθ=√(1-cos²θ)=√[1-{2p/(p²+1)}²]=√[1-4p²/(p²+1)²]=√[{(p²+1)²-4p²}/(p²+1)²]=√[(p⁴+2p²+1-4p²)/(p²+1)²]=√(p⁴-2p²+1)/(p²+1)=√(p²-1)²/(p²+1)=(p²-1)/(p²+1)∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1) Ans. |
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