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
New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
on 2011-11-27 01:59:42 (