Question

If sec x + tan x= k then prove that sin x =k^2-1/k^2+1

Answer 1

Given secx + tanx = k ...........1
wkt from 2nd identity
sec^2x-tan^2x = 1
a^2-b^2= (a+b) (a-b)
(secx + tanx) (secx - tanx)=1
(k) (secx - tanx) =1 from eq 1
secx-tanx = 1/k............2
adding equations 1 and 2, we get
secx + tanx + secx - tanx =1+1/k
secx + secx =1+ 1/ k
2secx=k^2+1/k
secx = k^2+1/2k
we know that secx = 1/cosx
cosx=1/k^2+1/2k
cosx = 2k/k^2+1
we know that from 1st identity
sinx = √(1-cosx^2)
sinx= √(1-(2k/k^2+1)^2
sinx = √(1-(4k^2/(k^2+1)^2
sinx = √(k^2+1)^2-4k^2)/k^2+1
sinx = √(k^2-1)^2/(k^2+1)^2
sinx = k^2 - 1/k^2+1
hence proved