CBSE Class 11-science Answered
3tanx+cotx=5cosecx ,x is greater than 0 and less than 90. find value of x
Asked by | 28 Oct, 2012, 07:18: PM
Expert Answer
3tanx+cotx=5cosecx
Multiply both sides with sinx* cosx
3(sinx)^2 + (cosx)^2 = 5cosx
We know (sinx)^2=1-(cosx)^2
3-3*(cosx)^2 + (cosx)^2=5cosx
let cosx=t
3-3t^2 + t^2=5t
0=2(t^2 )+5t-3
2(t^2 )+5t-3=0
(2t-1)(t+3)=0
so
2t-1=0 or t+3=0
so t=1/2 or t=-3
but t= cosx > -1, so t is not -3
Hence t=1/2
That is, cosx=1/2
Therefore x=60degrees
We know (sinx)^2=1-(cosx)^2
3-3*(cosx)^2 + (cosx)^2=5cosx
let cosx=t
3-3t^2 + t^2=5t
0=2(t^2 )+5t-3
2(t^2 )+5t-3=0
(2t-1)(t+3)=0
so
2t-1=0 or t+3=0
so t=1/2 or t=-3
but t= cosx > -1, so t is not -3
Hence t=1/2
That is, cosx=1/2
Therefore x=60degrees
Answered by | 28 Oct, 2012, 08:30: PM
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