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# Recent problems solved by 'Aswathy'

 Trigonometry-basics/535932: How do I verify these two identities? Here are the two identities. 1. (sin(x)sec^2(x) + sin(x) - 2tan(x)) / (cos(x) - 1)^2 = sec(x)tan(x) 2. (sin^3(x)-cos^3(x)) / (sin(x) - cos(x)) = 1 + sin(x)cos(x)1 solutions Answer 352689 by Aswathy(23)   on 2011-11-27 01:59:42 (Show Source): You can put this solution on YOUR website!(1) LHS =[sinx sec^2x + sinx - 2tanx] / (cosx - 1)^2 (tanx=sinx/cosx=secx/cosecx) =[sinx(sec^2x + 1- 2secx)]/[(1/secx-1]^2 =[sinx(sec^2x + 1- 2secx)] / [(1-secx)/secx]^2 =[sinx(sec^2x + 1- 2secx)] / [(1-2secx+ sec^2x)/sec^2x] =[sinx(sec^2x + 1- 2secx)] =sinx*sec^2x [cancelling out (sec^2x + 1- 2secx)] =sinx*1/cosx*secx (tanx=sinx/cosx=secx/cosecx) =tanx*secx=secx*tanx=RHS (2) LHS =[sin^3(x)-cos^3(x)] /[sin(x) - cos(x)] =[(sinx-cosx)(sin^2x+sinxcosx+cos^2x)] / (sinx-cosx) =sin^2+sinxcosx+cos^2x {cancelling out (sinx-cosx)] =1+sinxcosx (using identity sin^2+cos^2x=1) =RHS
 Trigonometry-basics/529961: prove that left side equation is equal to the right side 1+cosx/1-cosx + 1-sinx/1+sinx = 2(sinx + secx)/tanx-sinx+secx 1 solutions Answer 350167 by Aswathy(23)   on 2011-11-13 01:32:31 (Show Source): You can put this solution on YOUR website!Please write this question using brackets so that everyone could understand.
 Trigonometry-basics/524900: The first three terms of an arithmetic progression are tan x, cos x, and sec x, respectively. If the k th term is cot x, find k.1 solutions Answer 348896 by Aswathy(23)   on 2011-11-07 07:12:13 (Show Source): You can put this solution on YOUR website!Note that AP means arithmetic progression. Here, a1=tanx, a2=cosx and a3=secx Common difference (d)of the given AP=a2-a1=cosx-tanx=cosx-(sinx/cosx) =(cos^2x-sinx)/(cosx)..............(1) Similarly,a3-a2=secx-cosx=(1/cosx)-cosx=(1-cos^2x)/cosx=(sin^2x)/cosx........(2) Since the given terms are in AP,then the difference between the terms will be equal. Therefore,a2-a1=a3-a2 then,cosx-tanx=secx-cosx cosx+cosx=secx+tanx 2cosx=(1/cosx)+(sinx/cosx) 2cosx=(1+sinx)/cosx 2cos^2x=1+sinx.................(3) Similarly,now from (1) and(2), (cos^2x-sinx)/(cosx)=(sin^2x)/(cosx) then,cos^2x-sinx=sin^2x {cancelling out cosx on both sides of the equation} cos^2x=sinx+sin^2x cos^2x=sinx(1+sinx) cos^2x=sinx(2cos^2x) [from (3)] cos^2x/2cos^2x=sinx 1/2=sinx {cancelling out cos^2x on both sides of the equation} Therefore,x=30 as sinx=1/2 Let ak(k is subscript of a) be the kth term which is cotx here. In the given AP,d=a2-a1=cos30-tan30=1/2sqrt(3) Now,kth term ,ak=a1+(k-1)d then,cos30=tan30+(k-1)2/2sqrt(3) By solving this equation we get,k=5 which is our required answer. So,k=5
 Trigonometry-basics/522702: (tan(theta)-cot(theta))/(tan(theta)+cot(theta))+2cos^2(theta)=11 solutions Answer 348355 by Aswathy(23)   on 2011-11-04 10:09:24 (Show Source): You can put this solution on YOUR website!Note that I have typed angle A instead of theta. LHS= (tan A-cot A)/(tan A+cot A)+2cos^2 A = [(sin A/cos A)-( cos A/sin A)] / [(sin A/cos A)+( cos A/sin A)] +2cos^2 A = [(sin^2 A-cos^2 A)/(sin A*cos A)] /[( sin^2 A+cos^2 A)]/( sin A*cos A) +2cos^2 A = (sin^2 A-cos^2 A)/1 +(2cos^2 A) [by using the identity sin^2 A+cos^2 A=1 and by canceling out the common term sin A*cos A] = sin^2 A-cos^2 A+2cos^2 A =sin^2 A+cos ^2 A =1=RHS
 Trigonometry-basics/519724: Prove the following is an identity (5 cos E - 2 sin E)^2 + (2 cos E + 5 sin E)^2 =291 solutions Answer 346478 by Aswathy(23)   on 2011-10-26 07:55:47 (Show Source): You can put this solution on YOUR website!First before checking out the answer just check out the RHS side of your question. LHS=(5cos E - 2sin E)^2 + (2cos E +5 sin E)^2 =25cos^2E-20cosEsinE + 4sin^2E + 4cos^2E +20cosEsinE+25sin^2E =1+0+1 (using identity cos^2E+sin^2E=1) =2 =RHS
 Trigonometry-basics/520098: Trying to prove this is an identity. cos^3 J sin^2 J = cos^3 J - cos^5 J1 solutions Answer 346457 by Aswathy(23)   on 2011-10-26 04:07:47 (Show Source): You can put this solution on YOUR website!LHS=cos^3J sin^2J =cos^3J (1-cos^2J) (by using the identity sin^2 J+cos^2 J=1) =cos^3 J-cos^5 J=RHS
 Trigonometry-basics/516659: Multiply and simplify (sin θ + cos θ) (sin θ + cos θ) - 1 / sin θ1 solutions Answer 344874 by Aswathy(23)   on 2011-10-19 08:15:41 (Show Source): You can put this solution on YOUR website!(sin θ + cos θ) (sin θ + cos θ) - 1 / sin θ =[(sinθ+cosθ)^2-1]/sinθ =[sin^2+2cosθsinθ+cos^2-1]/sinθ =[1-1+2cosθsinθ]/sinθ (using the identity sin^2θ+cos^2θ=1) =2cosθsinθ/sinθ =2sinθ (by canceling out common factors)
 Trigonometry-basics/516677: Rewrite over a common denominator sin Multiply and simplify (sin θ + cos θ / cos θ) + (cos θ - sin θ / sin θ)1 solutions Answer 344873 by Aswathy(23)   on 2011-10-19 07:59:54 (Show Source): You can put this solution on YOUR website!(sin θ + cos θ / cos θ) + (cos θ - sin θ / sin θ) =(sin^2θ+cosθsinθ+cos^2θ-sinθcosθ)/cosθsinθ =(1)/cosθsinθ =secθcosecθ
 Trigonometry-basics/512878: cosθ/sinθ=1/tanθ1 solutions Answer 344089 by Aswathy(23)   on 2011-10-16 00:28:38 (Show Source): You can put this solution on YOUR website!costheta/sintheta=cot theta=1/tan theta
 Square-cubic-other-roots/485853: What is 9 divided by the cubed root of 3?1 solutions Answer 332207 by Aswathy(23)   on 2011-08-24 23:22:41 (Show Source): You can put this solution on YOUR website!9/cube rt of 3=(9/cube rt of 3)whole cubed =729/3=243
 Age_Word_Problems/481084: "5 years ago mother was 7 times as old as her daughter. 5 years hence she will be 3 times as old as her daughter. find their present age?"1 solutions Answer 329447 by Aswathy(23)   on 2011-08-13 00:12:52 (Show Source): You can put this solution on YOUR website!Let daughter's and mother's present ages be x and y respectively. As,5 years ago mother's age was 7 times her daughter's age, so,mother's will be y-5=7(x-5) Similarly,5 years from now the mother's age will be y+5=3(x+5) By solving both the equations we get x=10 & y=40. So the present ages of mother and her daughter are 40 & 10 years respectively.
 Travel_Word_Problems/480464: A certain fish requires 3 hours to swim 15 kilometers downstream. The return trip upstream take 5 hours. Find the speed of the fish in still water. How fast is the current of the stream?1 solutions Answer 329044 by Aswathy(23)   on 2011-08-11 08:53:17 (Show Source): You can put this solution on YOUR website!Let x and y be the speed of fish in still water and water current respectively. Then the speed of fish (1)when it swims downstream, x+y=15/3=5 (v=s/t) (2)when it swims upstream, x-y=15/5=3 so by solving the above pair of linear equation we get x=4 & y=1. So, the speed of the fish in still water is 4km/h & the speed of the current is 1km/h.
 Quadratic_Equations/480470: x=? in Equation x^2+4=9 1 solutions Answer 329035 by Aswathy(23)   on 2011-08-11 08:07:26 (Show Source): You can put this solution on YOUR website!x^2+4=9 x^2=9-4 x^2=5 x=+/-sqrt(5)
 Equations/480443: The sum of three numbers is 33. The first number is 3 less than the second, while the third is 3 more than the second. What are the numbers?1 solutions Answer 329031 by Aswathy(23)   on 2011-08-11 05:22:26 (Show Source): You can put this solution on YOUR website!Let x-3, x, x+3 be the 3 numbers. Then their sum, x-3+x+x+3=33 3x=33 x=11 x-3=8 x+3=14 The 3 numbers whose sum is 33 are 8, 11, and 14.
 Geometry_Word_Problems/479958: A lot is in the shape of a right triangle. The short leg measures 180 meters the hypotinus is 60 meters longer than the longer leg. How long is the longer leg?1 solutions Answer 328830 by Aswathy(23)   on 2011-08-10 02:50:38 (Show Source): You can put this solution on YOUR website!Let the measure of longer leg be x. Then, by using pythagoras theorem, x^2+(180)^2=(60+x)^2 x^2+32400=3600+120x+x^2 28800=120x x=240 so the longer leg is 240 meters long.
 Travel_Word_Problems/479986: The distance between two cities is 100 miles, and a woman drives from one city to the other at rate of 50 mph. At what rate must she return if the total travel time is three hours and forty minutes?1 solutions Answer 328828 by Aswathy(23)   on 2011-08-10 02:16:35 (Show Source): You can put this solution on YOUR website!The rate at which she had to return is 27.248mph
 Quadratic_Equations/479996: Subtract the polynomials (2w^2 +4w+4)(9w^2+4)1 solutions Answer 328825 by Aswathy(23)   on 2011-08-10 01:59:37 (Show Source): You can put this solution on YOUR website!(2w^2+4w+4)-(9w^2+4) =-7w^2+4w =-w(7w-4)
 Linear_Equations_And_Systems_Word_Problems/479623: sum of three consecutive odd no. is 63.find them1 solutions Answer 328600 by Aswathy(23)   on 2011-08-09 08:22:39 (Show Source): You can put this solution on YOUR website!Let 3 consecutive odd numbers be 2x+1,2x+3 and 2x+5. Then according to the given conditions, (2x+1)+(2x+3)+(2x+5)=63 6x+9=63 6x=54 x=54/6 x=9 2x+1=19 2x+3=21 2x+5=23 Thus the three consecutive odd numbers are 19,21 & 23.
 expressions/479619: (-x)^2-5x for x=-31 solutions Answer 328592 by Aswathy(23)   on 2011-08-09 07:38:51 (Show Source): You can put this solution on YOUR website!p(x)=(-x)^2-5x p(-3)=(-(-3)^2-5(-3) =9+15=24
 Geometry_proofs/478309: 1 Show that A is not between any two points of triangle ABC1 solutions Answer 328588 by Aswathy(23)   on 2011-08-09 06:06:12 (Show Source): You can put this solution on YOUR website!Let us assume that point A is between 2 points B & C. So A is on the line segment BC.So this means that ABC is no longer a triangle.This is because of our wrong assumption that A is between 2 points B and C. This leads to the conclusion that point A is not between any two points of a triangle ABC.
 Geometry_proofs/467155: Given: Line AB is parallel to Line CD; Ray BF bisects Angle ABE; Ray DG bisects Angle CDB PROVE: Line BF is parallel to Line DG1 solutions Answer 328587 by Aswathy(23)   on 2011-08-09 05:36:25 (Show Source): You can put this solution on YOUR website!As, AB is parallel to CD.Then by corresponding angles axiom, ABE=CDB as,FB & GD are angle bisectors So,1/2ABE=1/2CDB So,FBE=GDB Now by converse of corresponding angles axiom, BF is parallel to GD