Question

IF cot X =(1-n)cot(x-y), then prove that

tan y= n tan x/ (sec 2x - n tan 2x)

Answer 1

We know cotx = 1/tanx.
​​​​​....Using this equation...we re-arrange ....

tan(x-y) = (1-n)tanx
tan(x-y)=tanx - ntanx
. . or...on rearranging....
ntanx = tanx - tan(x-y)
we know tan(x-y) = (tanx - tany)/(1+ tanxtany)
.......we apply this formula in oru solutuon .....
we get....
ntanx = tanx - [(tanx -tany)/(1+tanxtany)]
....cross multiply and open the brackets in the RHS.....
we get
ntanx = [tanx(1+tanxtany) -tanx + tany ]/[1+ tanx tany]
.... on simplifing.....
ntanx = (tan^2xtany + tany)/(1+ tanxtany)
....taking tany as common from numerator....
ntanx = [(tan^2x + 1)tany/(1+tanxtany)]
.... we know tan^2x + 1 = sec^2x.....
ntanx = (sec^2x tany)/(1+ tanxtany). ......cross multiply and expand......
sec^2x tany = ntanx + ntan^2x tany
.. taking tany terms to one side.....
ntanx = tany(sec^2x - ntan^2x)
or... tany = ntanx/( sec^2x - ntan^2x).
.... hence proved....